Define the following terms:
(i) Angle
(ii) Interior of an angle
(iii) Obtuse angle
(iv) Reflex angle
(v) Complementary angles
(vi) Supplementary angles
(i) Angle – A shape formed by two lines or rays diverging from a common vertex.
Types of angle: (a) Acute angle (less than 90°)
(b) Right angle (exactly 90°)
(c) Obtuse angle (between 90° and 180°)
(d) Straight angle (exactly 180°)
(e) Reflex angle (between 180° and 360°)
(f) Full angle (exactly 360°)
(ii) Interior of an angle – The area between the rays that make up an angle and extending away from the vertex to infinity.
The interior angles of a triangle always add up to 180°.
(iii) Obtuse angle – It is an angle that measures between 90 to 180 degrees.
(iv) Reflex angle – It is an angle that measures between 180 to 360 degrees.
(v) Complementary angles – Two angles are called complementary angles if the sum of two angles is 90°.
(vi) Supplementary angles – Angles are said to be supplementary if the sum of two angles is 180°.
If = 36°27’46’’and =28°43’39’’, find +.
65°11’25’
+ = 36°27’46’’ + 28°43’39’’
= 64°70’85’’
∵ 60’ = 1° ⇒ 70’ = 1°10’
60’’ = 1’ ⇒ 85’’ = 1’ 25’’
∵ + = 65°11’25’’
Find the difference between two angles measuring 36° and 24°28’30’’
11°31’30’’
36° - 24°28’30’’ = 35°59’60’’ - 24°28’30’’
= 11°31’30’’
Find the complement of each of the following angles.
(i) 58°
(ii) 16°
(iii) of a right angle
(iv) 46°30’
(v) 52°43’20’’
(vi) 68°35’45’’
(i) 32°
Complement of angle = 90° – θ
Complement of 58° = 90° - 58°
= 32°
(ii) 74°
Complement of angle = 90° – θ
Complement of 58° = 90° - 16°
= 74°
(iii) 30°
Right angle = 90°
of a right angle = × 90°
= 60°
Complement of 60° = 90° - 60°
= 30°
(iv) 43°30’
Complement of angle = 90° – θ
Complement of 46°30’ = 90° - 46°30’
= 89°60’ - 46°30’
(v) 37°16’40’’
Complement of angle = 90° – θ
Complement of 52°43’20’’ = 90° - 52°43’20’’
= 89°59’60’’ - 52°43’20’’
= 37°16’40’’
(vi) 21°24’15’’
Complement of angle = 90° – θ
Complement of 68°35’45’’ = 90° - 68°35’45’’
= 89°59’60’’ - 68°35’45’’
= 68°35’45’’
Find the supplement of each of the following angles.
(i) 63°
(ii) 138°
(iii) of a right angle
(iv) 75°36’
(v) 124°20’40’’
(vi) 108°48’32’’
(i) 117°
Supplement of angle = 180° – θ
Supplement of 58° = 180° - 63°
= 117°
(ii) 42°
Supplement of angle = 180° – θ
Supplement of 58° = 180° - 138°
= 42°
(iii) 126°
Right angle = 90°
of a right angle = × 90°
= 54°
Supplement of 54° = 180° - 54°
= 126°
(iv) 104°24’
Supplement of angle = 180° – θ
Supplement of 75°36’ = 180° - 75°36’
= 179°60’ - 75°36’
= 104°24’
(v) 55°39’20’’
Supplement of angle = 180° – θ
Supplement of 124°20’40’ = 180° - 124°20’40’’
= 179°59’60” - 124°20’40’’
= 55°39’20’’
(vi) 71°11’28’’
Supplement of angle = 180° – θ
Supplement of 108°48’32’’ = 180° - 108°48’32’’
= 179°59’60” - 108°48’32’’
= 71°11’28’’
Find the measure of an angle which is
(i) equal to its complement,
(ii) equal to its supplement.
(i) 45°
Let, measure of an angle = X
Complement of X = 90° – X
Hence,
⇒ X = 90° – X
⇒ 2X = 90°
⇒ X = 45°
Therefore measure of an angle = 45°
(ii) 90°
Let, measure of an angle = X
Supplement of X = 180° – X
Hence,
⇒ X = 180° – X
⇒ 2X = 180°
⇒ X = 90°
Therefore measure of an angle = 90°
Find the measure of an angle which is 36° more than its complement.
63°
Let, measure of an angle = X
Complement of X = 90° – X
According to question,
⇒ X = (90° – X) + 36°
⇒ X + X = 90° + 36°
⇒ 2X = 126°
⇒ X = 63°
Therefore measure of an angle = 63°
Find the measure of an angle which 25°less than its supplement.
(77.5)°
Let, measure of an angle = X
Supplement of X = 180° – X
According to question,
⇒ X = (180° – X) - 25°
⇒ X + X = 180° - 25°
⇒ 2X = 155°
⇒ X = (77.5)°
Therefore measure of an angle = (77.5)°
Find the angle which is four times its complement.
72°
Let the angle = X
Complement of X = 90° – X
According to question,
⇒ X = 4(90° – X)
⇒ X = 360° - 4X
⇒ X + 4X = 360°
⇒ 5X = 360°
⇒ X = 72°
Therefore angle = 72°
Find the angle which is five times its supplement.
150°
Let the angle = X
Supplement of X = 180° – X
According to question,
⇒ X = 5(180° – X)
⇒ X = 900° - 4X
⇒ X + 5X = 900°
⇒ 6X = 900°
⇒ X = 150°
Therefore angle = 150°
Find the angle whose supplement is four times its complement.
60°
Let the angle = X
Complement of X = 90° – X
Supplement of X = 180° – X
According to question,
⇒ 180° – X = 4(90° – X)
⇒ 180° – X = 360° – 4X
⇒ – X + 4X = 360° – 180°
⇒ 3X = 180°
⇒ X = 60°
Therefore angle = 60°
Find the angle whose complement is four times its supplement.
180°
Let the angle = X
Complement of X = 90° – X
Supplement of X = 180° – X
According to question,
⇒ 90° – X = 4(180° – X)
⇒ 180° – X = 720° – 4X
⇒ – X + 4X = 720° – 180°
⇒ 3X = 540°
⇒ X = 180°
Therefore angle = 180°
Two supplementary angles are in the ratio 3:2 Find the angles.
108°, 72°
Let angle = X
Supplementary of X = 180° – X
According to question,
X : 180° – X = 3 : 2
⇒ X / (180° – X) = 3 / 2
⇒ 2X = 3(180° – X)
⇒ 2X = 540° – 3X
⇒ 2X + 3X = 540°
⇒ 5X = 540°
⇒ X = 108°
Therefore angle = 108°
And its supplement = 180° – 108° = 72°
Two complementary angles are in the ratio 4:5 Find the angles.
40°, 50°
Let angle = X
Complementary of X = 90° – X
According to question,
X : 90° – X = 4 : 5
⇒ X / (90° – X) = 4 / 5
⇒ 5X = 4(90° – X)
⇒ 5X = 360° – 4X
⇒ 5X + 4X = 360°
⇒ 9X = 360°
⇒ X = 40°
Therefore angle = 40°
And its supplement = 90° – 40° = 50°
Find the measure of an angle, if seven times its complement is 10° less than three times its supplement.
25°
Let the measure of an angle = X
Complement of X = 90° – X
Supplement of X = 180° – X
According to question,
⇒ 7(90° – X) = 3(180° – X) - 10°
⇒ 630° – 7X = 540° – 3X - 10°
⇒ – 7X + 3X = 540° – 10° – 630°
⇒ - 4X = 100°
⇒ X = 25°
Therefore measure of an angle = 25°
In the adjoining figure, AOB is a straight line. Find the value of x.
118°
AOB is a straight line
Therefore, ∠AOB = 180°
⇒ ∠AOC + ∠BOC = 180°
⇒ 62° + x = 180°
⇒ x = 180° – 62°
= 118°
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, Find And
X=27.5, =77.5°=47.5°
AOB is a straight line
Therefore, + ∠COD + = 180°
⇒ (3x - 5)° + 55° + (x + 20)° = 180°
⇒ 3x - 5° + 55° + x + 20° = 180°
⇒ 4x = 180° - 70°
⇒ 4x = 110°
⇒ x = 27.5°
= (3x - 5)°
= 3×27.5 – 5 = 77.5°
= (x + 20)°
= 27.5 + 20 = 47.5°
In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find,and .
X=32, =103°, ∠COD =45°=32°
AOB is a straight line
Therefore, + ∠COD + = 180°
⇒ (3x + 7)° + (2x - 19)° + x° = 180°
⇒ 3x + 7° + 2x - 19° + x° = 180°
⇒ 6x = 180° + 12°
⇒ 6x = 192°
⇒ x = 32°
= (3x + 7)°
= 3×32° + 7 = 103°
∠COD = (2x - 19)°
= 2×32° – 19 = 45°
= x
= 32°
In the adjoining figure, x: y: z =5:4:6. If XOY is a straight line, find the values of x, y and z
X=60, Y=48, Z=72
AOB is a straight line
Therefore, ∠XOP + ∠POQ + ∠YOQ = 180°
Given, x: y: z =5: 4: 6
Let ∠XOP = x° = 5a, ∠POQ = y° = 4a, ∠YOQ = z° = 6a
⇒ 5a + 4a + 6a = 180°
⇒ 15a = 180°
⇒ a = 12°
Therefore,
x = 5a = 5×12° = 60°
y = 4a = 4×12° = 48°
z = 6a = 6×12° = 72°
In the adjoining figure, what value of x will make AOB, a straight line?
X=28°
AOB is a straight line
Therefore, ∠AOB = 180°
⇒ (3x +20)° + (4x -36)° = 180°
⇒ 3x + 20° + 4x - 36° = 180°
⇒ 7x - 16° = 180°
⇒ 7x = 196°
⇒ x = 28°
Two lines AB and CD intersect at O. If =50°, find and .
=130°,=50°, =130°
Given AB and CD intersect a O
Therefore, = _____________________ (i)
And ∠BOC = ∠AOD _____________________ (ii)
=50°
Therefore, =50° from equation (i)
AOB is a straight line,
⇒ + = 180°
⇒ 50° + = 180°
⇒ = 180° - 50°
⇒ = 130°
∠AOD = ∠BOC = 130° from equation (ii)
In the adjoining figure, there coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.
X=4, Y=4, Z=50, t=90
Given, coplanar lines AB, CD and EF intersect at a point O.
Therefore, ∠AOF = ∠BOE ________________________ (i)
∠BOD = ∠AOC ________________________ (ii)
∠DOF = ∠COE ________________________ (iii)
x = y from equation (i)
t = 90 from equation (ii)
z = 50 from equation (iii)
∠AOF + ∠DOF + ∠BOD = 180° (from AOB straight line)
⇒ x + 50° + 90° = 180°
⇒ x = 180° - 140°
⇒ x = 40°
x = y = 40° from equation (i)
In the adjoining, there coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence, find,and .
)
∠AOD + ∠DOF + ∠BOF + ∠BOC + ∠COE + ∠AOE = 360°
⇒ 2x + 5x + 3x + 2x + 5x + 3x = 360°
⇒ 20x = 360°
⇒ x = 18°
∠AOD = 2x = 2× 18° = 36°
∠COE = 3x = 3× 18° = 54°
∠AOE = 4x = 4× 18° = 72°
Two adjacent angles on a straight line are in the ratio 5:4 Find the measure of each one of these angles.
100°, 80°
Explanation:
EOF is a straight line and its adjacent angles are ∠EOB and ∠FOB.
Let ∠EOB = 5a, and ∠FOB = 4a
∠EOB + ∠FOB = 180° (EOF is a straight line)
⇒5a + 4a = 180°
⇒9a = 180°
⇒ a = 20°
Therefore, ∠EOB = 5a
= 5 × 20° = 100°
And ∠FOB = 4a
= 4 × 20° = 80°
If two straight lines intersect each other in such a way that one of the angles formed measure 90°, show that each of the remaining angles measures 90°.
Proof
Given lines AB and CD intersect each other at point O and ∠AOC = 90°
∠AOC = ∠BOD (Opposite angles)
Therefore, ∠BOD = 90°
⇒ ∠BOD + ∠AOC = 180°
⇒ ∠BOC + 90° = 180°
⇒ ∠BOC = 90°
Now, ∠AOD = ∠BOC (Opposite angles)
Therefore,
∠AOD = 90°
Proved each of the remaining angles measures 90°.
Two lines AB and CD intersect at a point O such that + =280°, as shown in the figure. Find all the four angles.
=140°,=40°, = 140°, ∠BOD = 40°
Given lines AB and Cd intersect at a point O and + =280°
= (Opposite angle)
⇒ + = 280°
⇒ + = 280°
⇒ 2 = 280°
⇒ = 140°
= = 140°
Now,
+ = 180° (Because AOB is a straight line)
⇒ + 140° = 180°
⇒ = 40°
= ∠BOD = 40°
In the given figure, ray OC is the bisector of and OD is the ray opposite to OC. Show that =.
Proof
Given OC is the bisector of
Therefore, ∠AOC = ∠COB ______________________ (i)
DOC is a straight line,
+= 180° _______________ (ii)
Similarly, += 180° ________________ (iii)
From equations (i) and (ii)
⇒ += +
⇒+ = + (from equation (i))
⇒ = Proved
In the given figure, AB is a mirror; PQ is the incident ray and QR, the reflected ray. If =112°, Find .
34°
Angle of incidence =angle of reflection.
Therefore, ∠PQA = ∠BQR _____________________ (i)
⇒ ∠BQR + ∠PQR + ∠PQA = 180°[Because AQB is a straight line]
⇒ ∠BQR + 112° + ∠PQA = 180°
⇒ ∠BQR + ∠PQA = 180° - 112°
⇒ ∠PQA + ∠PQA = 68° [from equation (i)]
⇒ 2 ∠PQA = 68°
⇒ ∠PQA = 34°
If two straight lines intersect each other then prove that the ray opposite to the bisector of one of the angles so formed bisects the vertically opposite angle.
Given, lines AB and CD intersect each other at point O.
OE is the bisector of ∠ BOD.
TO prove: OF bisects ∠AOC.
Proof:
AB and CD intersect each other at point O.
Therefore, ∠ AOC = ∠ BOD
∠ 1 = ∠ 2 [OE is the bisector of ∠BOD] _________________ (i)
∠ 1 = ∠ 3 and ∠ 2 = ∠ 4 [Opposite angles] ____________ (ii)
From equations (i) and (ii)
∠ 3 = ∠ 4
Hence, OF is the bisector of ∠AOC.
Prove that the bisectors of two adjacent supplementary angles include a right angle.
Given, ∠AOC and ∠BOC are supplementary angles
OE is the bisector of ∠BOC and
OD is the bisector of ∠AOC
Therefore, ∠1 = ∠2 and ∠3 = ∠ 4 ________________ (i)
∠BOC + ∠AOC = 180° [Because AOB is a straight line]
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠1 + ∠3 + ∠3 = 180° [From equation (i)]
⇒ 2 ∠1 +2 ∠3 = 180°
⇒ 2(∠1 + ∠3) = 180°
⇒ ∠1 + ∠3 = 90°
Hence, ∠EOD = 90° proved.
In the adjoining figure, AB ||CD are cut by a transversal t at E and F respectively. If =70°, Find measure of each of the remaining marked angles.
=110°,=70°,=110°, =70°,=110°,=70°, =110°
Given AB ||CD are cut by a transversal t at E and F respectively.
And ∠1 = 70°
∠1 = ∠3 = 70° [Opposite angles]
∠5 = ∠1 = 70° [Corresponding angles]
∠3 = ∠7 = 70° [Corresponding angles]
∠1 + ∠2 = 180° [Because AB is a straight line]
⇒ 70° + ∠2 = 180°
⇒ ∠2 = 110°
∠4 = ∠2 = 110° [Opposite angles]
∠6 = ∠2 = 110° [Corresponding angles]
∠8 = ∠4 = 110° [Corresponding angles]
In the adjoining figure, AB ||CD are cut by a transversal t at E and F respectively. If =5:4, Find measure of each of the remaining marked angles.
=100°, =80°, =100°=80°, =100°, =80°, =100°, =80°
Given AB ||CD are cut by a transversal t at E and F respectively.
And = 5:4
Let ∠1 = 5a and ∠2 = 4a
∠1 + ∠2 = 180° [Because AB is a straight line]
⇒ 5a + 4a = 180°
⇒ 9a = 180°
⇒ a = 20°
Therefore, ∠1 = 5a
∠1 = 5 × 20° = 100°
∠2 = 4a
∠2 = 4 × 20° = 80°
∠3 = ∠1 = 100° [Opposite angles]
∠4 = ∠2 = 80° [Opposite angles]
∠5 = ∠1 = 100° [Crossponding angles]
∠6 = ∠4 = 80° [Crossponding angles]
∠7 = ∠5 = 100° [Opposite angles]
∠8 = ∠6 = 80° [Opposite angles]
In the adjoining figure, ABCD is a quadrilateral in which AB||DC and AD||BC. Prove that =.
Given AB||DC and AD||BC
Therefore, +=180° _____________ (i)
+=180° _____________ (ii)
From equations (i) and (ii)
+ = +
= Proved.
In each of the figure given below, AB||CD. Find the value of x in each case.
(i)
(ii)
(iii)
(i) x = 100
Given AB||CD, ∠ABE = 35° and ∠EDC = 65°
Draw a line PEQ||AB or CD
∠1 = ∠ABE = 35°[AB||PQ and alternate angle] _______________ (i)
∠2 = ∠EDC = 65°[CD||PQ and alternate angle] _______________ (ii)
From equations (i) and (ii)
∠1 + ∠2 = 100°
⇒ x = 100°
(ii) x=280
Given AB||CD, ∠ABE = 35° and ∠EDC = 65°
Draw a line POQ||AB or CD
∠1 = ∠ABO = 55°[AB||PQ and alternate angle] _______________ (i)
∠2 = ∠CDO = 25°[CD||PQ and alternate angle] _______________ (ii)
From equations (i) and (ii)
∠1 + ∠2 = 80°
Now,
∠BOD + ∠DOB = 360°
⇒ 80° + x° = 360°
⇒ x = 280°
(iii) x=120
Given AB||CD, ∠BAE = 116° and ∠DCE = 124°
Draw a line EF||AB or CD
∠BAE + ∠PAE = 180° [Because PAB is a straight line]
⇒ 116° + ∠3 = 180°
⇒ ∠3 = 180° - 116°
⇒ ∠3 = 64°
Therefore,
∠1 = ∠3 = 64° [Alternate angles] ____________________ (i)
Similarly, ∠4 = 180° - 124°
∠4 = 56°
Therefore,
∠2 = ∠4 = 56° [Alternate angles] ____________________ (ii)
From equations (i) and (ii)
⇒ ∠1 + ∠2 = 64° + 56°
⇒ x = 120°
In the given figure, AB||CD ||EF. Find the value of x.
X=20
Given AB||CD||EF, ∠ABC = 70° and ∠CEF = 130°
AB||CD
Therefore,
∠ABC = ∠BCD = 70° [Alternate angles] ________________ (i)
EF||CD
Therefore,
∠DCE + ∠CEF = 180°
⇒ ∠DCE + 130° = 180°
⇒ ∠DCE = 50°
Now,
∠BCE + ∠DCE = ∠BCD
⇒ x + 50° = 70°
⇒ x = 20°
In the given figure, AB||CD. Find the value of x.
X=110
Given AB||CD, ∠DCE = 130° and ∠AEC = 20°
Draw a line EF||AB||CD
CD||EF
Therefore, ∠DCE + ∠CEF = 180°
⇒ 130° + ∠1 = 180°
⇒ ∠1 = 180° - 130°
⇒ ∠1 = 50°
AB||EF
Therefore, ∠BAE + ∠AEF = 180°
⇒ x + ∠1 + 20° = 180°
⇒ x + 50° + 20° = 180°
⇒ x = 180° - 70°
⇒ x = 110°
In the given figure, AB||CD. Prove that -=.
Draw a line EF||AB||CD.
+=180° [Because AB||EF and AE is the transversal] __________________ (i)
+=180° [Because DC||EF and CE is the transversal] __________________ (ii)
From equations (i) and (ii)
⇒ +=+
⇒ -=-
⇒ - = Proved.
In the given figure, AB||CD and BC||ED. Find the value of x.
X=105
Given AB||CD and BC||ED.
AB||CD
Therefore, ∠BCF = ∠EDC = 75° [Crossponding angles]
∠ABC + ∠BCF = 180° [Because AB||DCF]
⇒ x + 75° = 180°
⇒ x = 105°
In the given figure, AB||CD. Prove that p+q-r=180
Given AB||CD, ∠AEF = P°, ∠EFG = q°, ∠FGD = r°
Draw a line FH||AB||CD
∠HFG = ∠FGD = r° [Because HF||CD and alternate angles] ___________ (i)
∠EFH = ∠EFG - ∠HFG
⇒ ∠EFH = q – r ______________________ (i)
∠AEF + ∠EFH = 180° [Because AB||HF]
⇒ ∠AEF + ∠EFH = 180°
⇒ p + (q – r) = 180°
⇒ p + q – r = 180°Proved.
In the given figure, AB||PQ. Find the value of x and y.
x=70, y=50
Given AB||PQ
∠GEF + 20° + 75° = 180°[Because EF is a straight line]
⇒ ∠GEF = 180° - 95°
⇒ ∠GEF = 85°________________ (i)
In triangle EFG,
⇒ X + 25° + 85° = 180° [∠GEF = 85°]
⇒ X = 60°
Now,
⇒ ∠BEF + ∠EFQ = 180°[Interior angles on same side of transversal]
⇒ (20° + 85°) + (25° + Y) = 180°
⇒ Y = 180° - 130°
⇒ Y = 50°
In the given figure, AB ||CD. Find the value of x.
Given AB||CD
Therefore, ∠BAC + ∠ACD = 180°
⇒ 75° + ∠ACD = 180°
⇒ ∠ACD = 105°_________________ (i)
∠ACD = ∠ECF = 105°[Opposite angles]
In triangle CEF,
⇒ ∠CEF + ∠EFC + ∠FCE = 180°
⇒ x + 30° + 105° = 180°
⇒ x + 135° = 180°
⇒ x = 45°
In the given figure, AB||CD. Find the value of x.
x=20
Given AB||CD
Therefore,
= [Crossponding angles]
= 95° ___________________ (i)
In CD straight line,
⇒ +=180°
⇒ 115° + = 180°
⇒ = 65°
In triangle GHQ,
⇒ ∠QGH + ∠GHQ + ∠GQH = 180°
⇒ 95° + 65° + x = 180°
⇒ x = 20°
In the given figure, AB||CD. Find the value of x, y and z.
Z=75, x=35, y=70
Given AB||CD
Therefore,
X = 35°[Alternate angles]
In triangle AOB,
⇒ x + 75° + y = 180°
⇒ 35° + 75° + y = 180°
⇒ y = 70°
⇒ ∠COD = y = 70°[Opposite angles]
In triangle COD,
⇒ z + 35° + ∠COD = 180°
⇒ z + 35° + 70° = 180°
⇒ z = 75°
In the given figure, AB||CD. Find the values of x, y and z.
x=105, y=75, z=50
Given AB||CD
Therefore,
⇒ ∠AEF = ∠EFG = 75°[Alternate angles]
⇒ y = 75°
For CD straight line,
⇒ x + y = 180°
⇒ x + 75° = 180°
⇒ x = 105°
Again,
⇒ ∠EGF + 125° = 180°
⇒ ∠EGF = 155°
In triangle EFG,
⇒ y + z + ∠EGF = 180°
⇒ 75°+ z + 155°= 180°
⇒ z + 130°= 180°
⇒ z = 50°
In the given figure, AB||CD and EF||GH. Find the values of x, y, z and t.
X=60, y=60, z=70, t=70
Given AB||CD and EF||GH
x = 60° [Opposite angles]
y = x = 60°[Alternate angles]
∠PQS = ∠APR = 110°[Crossponding angles]
∠PQS = ∠PQR + y = 110°_____________ (i)
For AB straight line,
⇒ y + z + ∠PQR = 180°
⇒ z + 110° = 180°[From equation (i)]
⇒ z = 70°
AB||CD
Therefore,
t = z = 70°[Because alternate angles]
For what value of x will the lines I and m be parallel to each other?
(i).
(ii)
(i) x=30
Given l||m
Therefore,
3x – 20° = 2x + 10° [Crossponding angles]
⇒ 3x – 2x = 10° + 20°
⇒ x = 30°
(ii) x=25
Given l||m
Therefore,
(3x + 5)° + 4x° = 180°
⇒ 7x + 5° = 180°
⇒ 7x = 175°
⇒ x = 25°
If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
AB⊥PQ,
Therefore, ∠ABD = 90° _____________ (i)
CD⊥PQ,
Therefore, ∠CDQ = 90° _____________ (ii)
From equations (i0 and (ii)
∠ABD = ∠CDQ = 90°
Hence, AB||CD because Cross ponding angles are equal.
=56°
In ΔABC,
∠A + ∠B + ∠C = 180° [Sum of angles]
⇒ ∠A + 76° + 48° = 180°
⇒ ∠A + 124° = 180°
⇒ ∠A = 56°
The angles of a triangle are in the ratio 2:3:4. Find the angles.
40°, 60°, 80°
Let the angles of triangle are 2a, 3a and 4a.
Therefore,
2a + 3a + 4a = 180°[Sum of angles]
⇒ 9a = 180°
⇒ a = 20°
Angles of triangle are,
2a = 2 × 20° = 40°
3a = 3 × 20° = 60°
4a = 4 × 20° = 80°
In , if 3= 4=6, calculate, and .
=80°, =60°, =40°
Let 3= 4=6 = a
Therefore,
∠A = a/3, ∠B = a/4, ∠C = a/6 ____________________ (i)
∠A + ∠B + ∠C = 180° [Sum of angles]
⇒ a/3 + a/4 + a/6 = 180°
⇒ 9a/12 = 180°
⇒ a = 240°
⇒ ∠A = a/3 = 240° /3 = 80°
⇒ ∠B = a/4 = 240° /4 = 60°
⇒ ∠C = a/6 = 240° /6 = 40°
In , if +=108°and+ =130°,Find ,and.
=50°, =58°,=72°
Given,
+= 108° ___________________ (i)
+ = 130° ___________________ (ii)
We know that sum of angles of triangle = 180°
∠A + ∠B + ∠C = 180° [Sum of angles]
∠A + 130°= 180° [From equation (ii)]
⇒ ∠A = 50°
Value of ∠A = 50° put in equation (i),
+ = 108°
⇒ 50° + = 108°
⇒ = 58°
Value of ∠B = 58° put in equation (ii),
+ = 130°
⇒ 58° + = 130°
⇒ = 72°
In , if +=125°and+ =113°,Find ,and.
=67°, =41°, =89°
Given,
+= 125° ___________________ (i)
+ = 113° ___________________ (ii)
We know that sum of angles of triangle = 180°
∠A + ∠B + ∠C = 180° [Sum of angles]
∠A + 113°= 180° [From equation (ii)]
⇒ ∠A = 67°
Value of ∠A = 50° put in equation (i),
+ = 125°
⇒ 67° + = 108°
⇒ = 41°
Value of ∠B = 41° put in equation (ii),
+ = 130°
⇒ 41° + = 130°
⇒ = 89°
In, if -=42°and-=21°, Find,and.
=95°,=53°,=32°
Given,
-= 42° ___________________ (i)
-=21° ___________________ (ii)
= 42° + [From equation (i)] __________________ (iii)
= - 21° [From equation (ii)] __________________ (iv)
We know that sum of angles of triangle = 180°
∠P + ∠Q + ∠R = 180° [Sum of angles]
⇒ 42° + ∠Q + ∠Q + ∠Q - 21° = 180° [From equation (iii) and (iv)]
⇒ 3 ∠Q + 21°= 180°
⇒ 3 ∠Q = 159°
⇒ ∠Q = 53°
Value of ∠Q = 53° put in equation (iii),
= 42° +
⇒ = 42° + 53°
⇒ = 95°
Value of ∠Q = 53° put in equation (iv),
= - 21°
⇒ = 53° - 21°
⇒ = 32°
The sum of two angles of a triangle is 116° and their difference is 24°.Find the measure of each angle of the triangle.
70°, 46°, 64°
Let ∠P, ∠Q and ∠R are three angles of triangle PQR.
Now,
∠P + ∠Q = 116° ___________________ (i)
∠P - ∠Q = 24° ___________________ (i)
Adding equation (i) and (ii),
2 ∠P = 140°
⇒ ∠P = 70° ______________________ (iii)
Subtracting equation (i) and (ii),
2 ∠Q = 92°
⇒ ∠Q = 46° _______________________ (iv)
We know that sum of angles of triangle = 180°
∠P + ∠Q + ∠R = 180° [Sum of angles]
⇒ 70° + 46° + ∠R = 180° [From equation (iii) and (iv)]
⇒ ∠R = 64°
Of the three angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angle.
54°, 54°, 72°
Let ∠P, ∠Q and ∠R are three angles of triangle PQR,
And ∠P = ∠Q = a ________________ (i)
Then, ∠R = a + 18°________________ (ii)
We know that sum of angles of triangle = 180°
∠P + ∠Q + ∠R = 180° [Sum of angles]
⇒ a + a + a + 18°= 180° [From equation (i) and (ii)]
⇒ 3a= 162°
⇒ a= 54°
Therefore,
∠P = ∠Q =54° [from equation (i)]
∠R = 54° + 18° [from equation (i)]
= 72°
Of the three angles of a triangle, one is twice the smallest and mother one is thrice the smallest. Find the angle.
60°, 90°, 30°
Let ∠P, ∠Q and ∠R are three angles of triangle PQR,
And ∠P is the smallest angle.
Now,
∠Q = 2 ∠P ________________ (i)
∠R = 3 ∠P ________________ (ii)
We know that sum of angles of triangle = 180°
∠P + ∠Q + ∠R = 180° [Sum of angles]
⇒ ∠P + 2 ∠P + 3 ∠P = 180° [From equation (i) and (ii)]
⇒ 6 ∠P = 180°
⇒ ∠P = 30°
Therefore,
⇒ ∠Q = 2 ∠P = 60° [from equation (i)]
⇒ ∠R = 3 ∠P = 90° [from equation (ii)]
In a right-angled triangle, one of the acute measures 53°. Find the measure of each angle of the triangle.
53°, 37°, 90°
Let PQR be a right angle triangle.
Right angle at P, then
∠P = 90° and ∠Q = 53° ____________________________________ (i)
We know that sum of angles of triangle = 180°
∠P + ∠Q + ∠R = 180° [Sum of angles]
⇒ 90° + 53° + ∠R = 180° [From equation (i)]
⇒ ∠R = 37°
If one angle of a triangle is equal to the sum of the other two, show that the triangle is right angled.
Proof
Let PQR be a right angle triangle,
Now,
∠P = ∠Q + ∠R __________________ (i)
We know that sum of angles of triangle = 180°
∠P + ∠Q + ∠R = 180° [Sum of angles]
⇒ ∠P + ∠P = 180° [From equation (i)]
⇒ 2 ∠P = 180°
⇒ ∠P = 90°
Hence, PQR is a right angle triangle Proved.
A is right angled at A. If AL BC, prove that =.
proof
We know that the sum of two acute angles of a right triangle is 90°.
Therefore,
+=90°
= 90°-
= 90°-____________________ (i)
+=90°
= 90°-____________________ (ii)
From equation (i) and (ii),
= Proved.
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Proof
Let ABC be a triangle,
Now,
< +___________________ (i)
< +___________________ (ii)
< +___________________ (iii)
2 < ++ [From equation (i)]
2 < 180° [Sum of angles of triangle]
< 90° ________________ (a)
Similarly,
< 90°__________________ (b)
< 90°__________________ (c)
From equation (a), (b) and (c), each angle is less than 90°
Therefore triangle is an acute angled Proved.
If each angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse angled.
Proof
Let ABC be a triangle,
Now,
> +___________________ (i)
> +___________________ (ii)
> +___________________ (iii)
2 > ++ [From equation (i)]
2 > 180° [Sum of angles of triangle]
> 90° ________________ (a)
Similarly,
> 90°__________________ (b)
> 90°__________________ (c)
From equation (a), (b) and (c), each angle is less than 90°
Therefore triangle is an acute angled Proved.
In the given figure, side BC of ABC is produced to D. If =128° and =43°,Find and .
∠BAC = 85°, ∠ACB = 52°
Given, =128° and =43°
In triangle ABC,
∠ACB + ∠ACD = 180° [Because BCD is a straight line]
⇒ ∠ACB + 128° = 180°
⇒ ∠ACB = 52°
∠ABC + ∠ACB + ∠BAC = 180° [Sum of angles of triangle ABC]
⇒ 43° + 52° + ∠BAC = 180°
⇒ ∠BAC = 85°
In the given figure, the side BC of has been produced on both sides-on the left to D and on the right to E. If =106° and =118°, Find the measure of each angle of the triangle.
74°, 62°, 44°
Given, =106° and =118°
∠ABD + ∠ABC = 180° [Because DC is a straight line]
⇒ 106° + ∠ABC = 180°
⇒ ∠ABC = 74°_______________ (i)
∠ACB + ∠ACE = 180° [Because BE is a straight line]
⇒ ∠ACB + 118° = 180°
⇒ ∠ACB = 62°_______________ (ii)
Now, triangle ABC
∠ABC + ∠ACB + ∠BAC = 180° [Sum of angles of triangle]
⇒ 74° + 62° + ∠BAC = 180° [From equation (i) and (ii)]
⇒ ∠BAC = 44°
Calculate the value of x in each of the following figure.
(i).
(ii)
(iii)
(iv)
(v)
(vi)
(i) 50°
Given, ∠BAE = 110° and ∠ACD = 120°
∠ACB + ∠ACD = 180° [Because BD is a straight line]
⇒ ∠ACB + 120° = 180°
⇒ ∠ACB = 60°_______________ (i)
In triangle ABC,
∠BAE = ∠ABC + ∠ACB
⇒ 110° = x + 60°
⇒ x = 50°
(ii) 120°
In triangle ABC,
∠A + ∠B + ∠C = 180° [Sum of angles of triangle ABC]
⇒ 30° + 40° + ∠C = 180°
⇒ ∠C = 110°
∠BCA + ∠DCA = 180° [Because BD is a straight line]
⇒ 110° + ∠DCA = 180°
⇒ ∠DCA = 70°_________________ (i)
In triangle ECD,
∠AED = ∠ECD + ∠EDC
⇒ x = 70°+ 50°
⇒ x = 120°
(iii) 55°
Explanation:
∠BAC = ∠EAF = 60°[Opposite angles]
In triangle ABC,
∠ABC + ∠BAC = ∠ACD
⇒ X°+ 60°= 115°
⇒ X°= 55°
(iv) 75°
Given AB||CD
Therefore,
∠BAD = ∠EDC = 60°[Alternate angles]
In triangle CED,
∠C + ∠D + ∠E = 180°[Sum of angles of triangle]
⇒ 45° + 60° + x = 180°[∠EDC = 60°]
⇒ x = 75°
(v) 30°
Explanation:
In triangle ABC,
∠BAC + ∠BCA + ∠ABC = 180°[Sum of angles of triangle]
⇒ 40° + 90° + ∠ABC = 180°
⇒ ∠ABC = 50°________________ (i)
In triangle BDE,
∠BDE + ∠BED + ∠EBD = 180°[Sum of angles of triangle]
⇒ x° + 100° + 50° = 180°[∠EBD = ∠ABC = 50°]
⇒ x° = 30°
(vi) x=30
Explanation:
In triangle ABE,
∠BAE + ∠BEA + ∠ABE = 180°[Sum of angles of triangle]
⇒ 75° + ∠BEA + 65° = 180°
⇒ ∠BEA = 40°
∠BEA = ∠CED = 40°[Opposite angles]
In triangle CDE,
∠CDE + ∠CED + ∠ECD = 180°[Sum of angles of triangle]
⇒ x° + 40° + 110° = 180°
⇒ x° = 30°
Calculate the value of x in the given figure.
x=130
Explanation:
In triangle ACD,
∠3 = ∠1 + ∠C __________________ (i)
In triangle ABD,
∠4 = ∠2 + ∠B __________________ (ii)
Adding equation (i) and (ii),
∠3 + ∠4 = ∠1 + ∠C + ∠2 + ∠B
⇒ ∠BDC = (∠1 + ∠2) + ∠C + ∠B
⇒ x°= 55°+ 30°+ 45°
⇒ x°= 130°
In the given figure, AD divides in the ratio 1:3 and AD=DB. Determine the value of.
X=90
Explanation:
∠BAC + ∠CAE = 180°[Because BE is a straight line]
⇒ ∠BAC + 108° = 180°
⇒ ∠BAC = 72°
Now,
AD = DB
=
∠BAD = ( �)72°= 18°
∠DAC = ( �)72°= 54°
In triangle ABC,
∠A + ∠B + ∠C = 180°[Sum of angles of triangle]
⇒ 72° + 18° + x = 180°
⇒ x = 90°
If the side of a triangle are produced in order, Prove that the sum of the exterior angles so formed is equal to four right angles.
Proof
In triangle ABC,
=+ ________________ (i)
=+________________ (ii)
=+________________ (iii)
Adding equation (i), (ii) and (iii),
++= 2(++)
⇒ ++= 2(180°) [Sum of angles of triangle]
⇒ ++= 360°Proved.
In the adjoining figure, show that +++++=360°
Proof
In triangle BDF,
++ = 180°[Sum of angles of triangle] _______________ (i)
In triangle BDF,
++ = 180°[Sum of angles of triangle] _______________ (ii)
From equation (i) and (ii),
(++) + (++) = (180°+180°)
⇒ +++ ++ = 360°Proved.
In ABC the angle bisectors of and meet at O. If =70°,Find .
125°
Given, bisector of and meet at O.
If OB and OC are the bisector of and meet at point O .
Then,
= 90°+
⇒= 90°+ 70°
⇒= 125°
The sides AB and AC of ABC have been produced to D and E respectively. The bisectors of and meet at O. If =40° find .
70°
Given, bisector of and meet at O.
If OB and OC are the bisector of and meet at point O .
Then,
= 90°-
⇒= 90°- 40°
⇒= 70°
In the given figure, ABC is a triangle in which::=3:2:1 and AC CD. Find the measure of.
60°
Given, ::= 3:2:1 and AC CD
Let, ∠A = 3a
∠B = 2a
∠C = a
In triangle ABC,
∠A + ∠B + ∠C = 180°[Sum of angles of triangle]
⇒ 3a + 2a + a = 180°
⇒ 6a = 180°
⇒ a = 30°
Therefore, ∠C = a = 30°
Now,
∠ACB + ∠ACD + ∠ECD = 180°[Sum of angles of triangle]
⇒ 30° + 90° + ∠ECD = 180°
⇒ ∠ECD = 60°
In the given figure, AM BC and AN is the bisector of. Find the measure of.
17.5°
Given, AMBC and “AN” is the bisector of.
Therefore,
= (-)
⇒ = (65° -30°)
⇒ = 17.5°
State ‘True’ or ‘false’:
(i) A triangle can have two right angles.
(ii) A triangle cannot have two obtuse angles.
(iii) A triangle cannot have two acute angles.
(iv) A triangle can have each angle less than 60°.
(v) A triangle can have each angle equal to 60°.
(vi) There cannot be a triangle whose angles measure 10°, 80° and 100°.
(i) False
Because, sum of angles of triangle equal to 180°.In a triangle maximum one right angle.
(ii) True
Because, obtuse angle measures in 90° to 180° and we know that the sum of angles of triangle is equal to 180°.
(iii) False
Because, in an obtuse triangle is one with one obtuse angle and two acute angles.
(iv) False
If each angles of triangle is less than 180° then sum of angles of triangle are not equal to 180°.
Any triangle,
∠1 + ∠2 + ∠3 = 180°
(v) True
If value of angles of triangle is same then the each value is equal to 60°.
∠1 + ∠2 + ∠3 = 180°
⇒ ∠1 + ∠1 + ∠1 = 180°[∠1 = ∠2 = ∠3]
⇒ 3 ∠1 = 180°
⇒ ∠1 = 60°
(vi) True
We know that sum of angles of triangle is equal to 180°.
Sum of angles,
= 10° + 80° + 100°
= 190°
Therefore, angles measure in (10°, 80°, 100°) cannot be a triangle.
If two angles are complements of each other, then each angle is
A. an acute angle
B. an obtuse angle
C. a right angle
D. a reflex angle
If two angles are complements of each other, then each angle is an acute angle
An angle which measures more than 180° but less than 360°, is called
A. an acute angle
B. an obtuse angle
C. a straight angle
D. a reflex angle
An angle which measures more than 180o but less than 360o, is called a reflex angle.
The complement of 72°40’ is
A. 107°20’
B. 27°20’
C. 17°20’
D. 12°40’
As we know that sum of two complementary – angles is 90o.
So, x + y = 90o
72°40’ + y= 90
y = 90o – 72°40’
y = 17o20’
The supplement of 54°30’ is
A. 35°30’
B. 125°30’
C. 45°30’
D. 65°30’
As we know that sum of two supplementary – angles is 180o.
So, x + y = 180o
54°30’ + y= 180
y = 180o – 54°30’
y = 125o30’
The measure of an angle is five times its complement. The angle measures
A. 25°
B. 35°
C. 65°
D. 75°
As we know that sum of two complementary – angles is 90o.
So, x + y = 90o
According to question y =5x
x + 5x= 90
6x = 90o
x = 15o
y = 75o
Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The larger of the two measures
A. 72°
B. 54°
C. 63°
D. 36°
As we know that sum of two complementary – angles is 90o.
So, x + y = 90o
Let x be the common multiple.
According to question angles would be 2x and 3x.
2x + 3x= 90
5x = 90o
x = 18o
2x = 36o
3x = 54o
So, larger angle is 54o
Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOD = ?
A. 63°
B. 117°
C. 17°
D. 153°
As we know that sum of adjacent angle on a straight line is 180o.
∠BOD + ∠BOC = 180°
∠BOC = 180° – 63°
∠BOC = 117°
In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ?
A. 95°
B. 85°
C. 90°
D. 55°
As we know that sum of adjacent angle on a straight line is 180o.
In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC = ?
A. 40°
B. 60°
C. 80°
D. 100°
As we know that sum of adjacent angle on a straight line is 180o.
According to question,
,
4x + 5x = 180o
9x =180o
X =20o
In the given figure, AOB is a straight line. If ∠AOC (3x + 10)° and ∠BOC = (4x – 26)°, then ∠BOC = ?
A. 96°
B. 86°
C. 76°
D. 106°
As we know that sum of adjacent angle on a straight line is 180o.
According to question,
∠ AOC = (3x + 10)°
∠ BOC = (4x – 26)°
3x + 10 + 4x – 26 = 180o
7x – 16 =180o
7x =196o
X= 28o
∠ BOC = (4x – 26)°
∠ BOC = 112° – 26°
∠ BOC = 86°
In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x°, and ∠BOD = 3x°, then ∠COD = ?
A. 80°
B. 100°
C. 120°
D. 140°
As we know that sum of all angles on a straight line is 180°
∠ AOC + ∠COD + ∠BOD = 180°
In the given figure, AOB is a straight line. If ∠AOC = (3x – 10)°, ∠COD = 50° and ∠BOD = (x + 20)°, then ∠AOC = ?
A. 40°
B. 60°
C. 80°
D. 50°
As we know that sum of all angles on a straight line is 180o.
Which of the following statements is false?
A. Through a given point, only one straight line can be drawn.
B. Through two given points, it is possible to draw one and only one straight line.
C. Two straight lines can intersect only at one point.
D. A line segment can be produced to any desired length.
Through a given point, we can draw infinite number of lines.
An angle is one – fifth of its supplement. The measure of the angle is
A. 15°
B. 30°
C. 75°
D. 150°
Let x be the common multiple.
According to question,
y= 5x
As we know that sum of two supplementary – angles is 180o.
So, x + y = 180o
x + 5x= 180
6x = 180o
x = 30o
In the adjoining figure, AOB is a straight line. If x : y : z = 4 : 5 : 6, then y = ?
A. 60°
B. 80°
C. 48°
D. 72°
Let n be the common multiple
As we know that sum of all angles on a straight line is 180o.
4n + 5n + 6n =180o
15n = 180o
N = 12o
Y = 5n = 60o
In the given figure, straight lines AB and CD intersect at O. If ∠AOC = ϕ, ∠BOC = θ and θ = 3θ, then θ = ?
A. 30°
B. 40°
C. 45°
D. 60°
As we know that sum of all angles on a straight line is 180o.
According to question,
In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD = ?
A. 65°
B. 115°
C. 110°
D. 125°
AC and BD intersect at O.
As we know that sum of all angles on a straight line is 180o.
In the given figure AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, then ∠AQP = ?
A. 72°
B. 18°
C. 36°
D. 54°
Incident ray makes the same angle as reflected ray.
So,
In the given figure AB || CD. If ∠OAB = 124°, ∠OCD = 136°, then ∠AOC = ?
A. 80°
B. 90°
C. 100°
D. 110°
Draw a line EF such that EF || AB and EF || CD crossing point O.
FOC + OCD = 180o (Sum of consecutive interior angles is 180o)
FOC = 180 – 136 = 44o
EF || AB such that AO is traversal.
OAB + FOA = 180o(Sum of consecutive interior angles is 180o)
FOA = 180 – 124 = 56o
AOC = FOC + FOA
= 56 + 44
=100o
In the given figure AB || CD and O is a point joined with B and D, as shown in the figure such that ∠ABO = 35 and ∠CDO = 40°. Reflex ∠BOD = ?
A. 255°
B. 265°
C. 275°
D. 285°
Draw a line EF such that EF || AB and EF || CD crossing point O.
ABO + EOB = 180o(Sum of consecutive interior angles is 180o)
EOB = 180 – 35 = 145o
EF || AB such that AO is traversal.
CDO + EOD = 180o(Sum of consecutive interior angles is 180o)
EOD = 180 – 40 = 140o
BOD = EOB + EOD
= 145 + 140
= 285o
In the given figure, AB || CD. If ∠ABO = 130° and ∠OCD = 110°, then ∠BOC = ?
A. 50°
B. 60°
C. 70°
D. 80°
According to question,
AB || CD
AF || CD (AB is produced to F, CF is traversal)
DCF=BFC=110o
Now, BFC + BFO = 180o(Sum of angles of Linear pair is 180o)
BFO = 180o – 110o = 70o
Now in triangle BOF, we have
ABO = BFO + BOF
130 = 70 + BOF
BOF = 130 – 70 =60o
So, BOC = 60o
In the given figure, AB || CD. If ∠BAO = 60° and ∠OCD = 110°, then ∠AOC = ?
A. 70°
B. 60°
C. 50°
D. 40°
According to question,
AB || CD
AB || DF (DC is produced to F)
OCD=110o
FCD = 180 – 110 = 70o(linear pair)
Now in triangle FOC, we have
FOC + CFO + OCF = 180o
FOC + 60 + 70 = 180o
FOC = 180 – 130
=50o
So, AOC = 50o
In the given figure, AB || CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD = ?
A. 130°
B. 150°
C. 80°
D. 100°
From O, draw E such that OE || CD || AB.
OE || CD and OC is traversal.
So,
DCO + COE = 180 (co –interior angles)
x + COE = 180
COE = (180 – x)
Now, OE || AB and AO is the traversal.
BAO + AOE = 180 (co –interior angles)
BAO + AOC + COE = 180
100 + 30 + (180 – x) = 180
180 – x = 50
X = 180 – 50 = 130O
In the given figure, AB || CD. If ∠CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
A. 65°
B. 55°
C. 45°
D. 75°
AB || CD
BAC = DCF = 80o
ECF + DCF = 180o (linear pair of angles)
ECF =100o
Now in triangle CFE,
ECF + EFC + CEF = 180o
CEF = 180o – 100o – 25o
=55o
In the given figure, AB || CD. If ∠APQ = 70° and ∠PRD = 120°, then ∠QPR = ?
A. 50°
B. 60°
C. 40°
D. 35°
APQ =PQR =70o
Now, in triangle PQR, we have
PQR + PRQ + QPQ =180o
70 + 60 + QPQ =180o
QPQ =180o – 130o
=50o
In the given figure, x = ?
A. α + β – γ
B. α – β + γ
C. α + β + γ
D. α + γ – β
AC is produced to meet OB at D.
OEC = 180 –
So,BEC = 180 – (180 – ) =
Now, x = BEC + CBE (Exterior Angle)
= +
=
If 3∠A = 4∠B = 6∠C, then A : B : C = ?
A. 3:4:6
B. 4:3:2
C. 2:3:4
D. 6:4:3
Let say
A =x/3
B = x/4
C = x/6
A + B + C = 180
x/3 + x/4 + x/6 = 180
(4x + 3x + 2x)/12 = 180
9x/12 = 180
X= 240
A =x/3 = 240/3 = 80
B = x/4 = 240/4 = 60
C = x/6 = 240/6 = 40
So, A:B:C = 4:3:2
In ΔABC, if ∠A + ∠B = 125° and ∠A + ∠C = 113°, then ∠A = ?
A. (62.5°)
B. (56.5)°
C. 58°
D. 63°
A + B + C = 180
C = 180 – 125 = 55o
A + C =113o
A =113 – 55 =58o
In ΔABC, if ∠A – ∠B = 42° and ∠B – ∠C = 21°, then ∠B = ?
A. 95°
B. 53°
C. 32°
D. 63°
A = B + 42
C = B – 21
A + B + C = 180
B + 42 + B + B – 21 =180
3B + 21 = 180
3B = 159
B =
In ΔABC, side BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ?
A. 60°
B. 40°
C. 80°
D. 50°
ACD + ACB = 180 (Linear pair of angles)
ACB = 60o
ABC = 40o
As we know that
ACB + ACB + BAC = 180o
BAC = 180 – 60 – 40
=80o
Side BC of ΔABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?
A. 65°
B. 75°
C. 50°
D. 55°
ABD + ABC = 180 (Linear pair of angles)
ABC = 180o – 125o=55o
ACE + ACB = 180 (Linear pair of angles)
ACB = 180o – 130o=50o
As we know that
ACB + ABC + BAC = 180o
BAC = 180 – 55 – 50
=75o
In the given figure, ∠BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Then ∠AED = ?
A. 120°
B. 100°
C. 80°
D. 110°
ACB + ABC + BAC =180
ACB = 180 – 50 – 30 = 100o(Sum of angles of triangle is 180)
ACB + ACD = 180 (linear pair of angles)
ACD = 180 – 100 = 80o
In triangle ECD,
ECD + CDE + DEC = 180
DEC = 180 – 80 – 40
= 60o
DEC + AED = 180o(linear pair of angles)
AED = 180o – 60o
= 120o
In the given figure, ∠BAC = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?
A. 50°
B. 30°
C. 40°
D. 25°
In triangle AEF,
BED = EFA + EAF
EFA = 100 – 40 = 60o
CFD = EFA (vertical opposite angles)
= 60o
In triangle CFD, we have
CFD + FCD + CDF = 180o
CDF = 180o – 90o – 60o
= 30o
So, BDE = 30o
In the given figure, BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50°, then ∠BOC = ?
A. 130°
B. 100°
C. 115°
D. 120°
In ∆ABC,
∠A + ∠B + ∠C=180°
50° + ∠B + ∠C=180°
∠B + ∠C=180°−50°=130°
∠B = 65°
∠C = 65°
Now in ∆OBC,
∠OBC + ∠OCB + ∠BOC=180°
∠BOC = 180° – 65° (∠OBC + ∠OCB = 65 because O is bisector of ∠B and ∠C)
= 115°
In the given figure, AB || CD. If ∠EAB = 50° and ∠ECD = 60°, then ∠AEB = ?
A. 50°
B. 60°
C. 70°
D. 55°
AB || CD and BC is traversal.
So, ∠DCB = ∠ABC = 60o
Now in triangle AEB, we have
∠ABE + ∠BAE + ∠AEB =180o
∠AEB =180o – 60o – 50o
= 70o
In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?
A. 20°
B. 25°
C. 30°
D. 35°
In triangle AOB,
∠AOB =180o – 75o – 55o
= 50o
∠AOB = ∠COD = 50o(Opposite angles)
Now in triangle COD,
∠ODC =180o – 100o – 50o
= 30o
In a ΔABC its is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and CD ⊥ AC. Then ∠ECD = ?
A. 60°
B. 45°
C. 75°
D. 30°
As per question,
So,
∠A = 90o
∠B = 60o
∠C = 30o
∠ACB + ∠ACD + ∠ECD = 180o (sum of angles on straight line)
∠ECD = 180o – 90o – 30o
= 60o
In the given figure, AB || CD. If ∠ABO = 45° and ∠COD = 100° then ∠CDO = ?
A. 25°
B. 30°
C. 35°
D. 45°
∠BOA = 100o (Opposite pair of angles)
So,
∠BAO = 180o – 100o – 45o
=35o
∠BAO = ∠CDO =35o (Corresponding Angles)
In the given figure, AB || DC, ∠BAD = 90°, ∠CBD = 28° and ∠BCE = 65°. Then ∠ABD = ?
A. 32°
B. 37°
C. 43°
D. 53°
∠BCE = ∠ABC =65o (Alternate Angles)
∠ABC = ∠ABD + ∠DBC
65o = ∠ABD + 28o
∠ABD = 65 – 28
= 37o
For what value of x shall we have l || m?
A. x = 50
B. x = 70
C. x = 60
D. x = 45
X + 20 = 2x – 30(Corresponding Angles)
2x –x = 30 + 20
X = 50o
For what value of x shall we have l || m?
A. x = 35
B. x = 30
C. x = 25
D. x = 20
4x + 3x + 5 = 180o (Interior angles of same side of traversal)
7x + 5 = 180o
7x = 175
X = 25o
In the given figure, sides CB and BA of ΔABC have been produced to D and E respectively such that ∠ABD = 110° and ∠CAE = 135°. Then ∠ACB = ?
A. 35°
B. 45°
C. 55°
D. 65°
∠ABC = 180 – 110 = 700 (Linear pair of angles)
∠BAC = 180 – 135 = 450 (Linear pair of angles)
So,
In Triangle ABC, we have
∠ABC + ∠BAC + ∠ACB = 180o
∠ACB = 180 – 70 – 45 = 650
In ΔABC, BD ⊥ AC, ∠CAE = 30° and ∠CBD = 40°. Then ∠AEB = ?
A. 35°
B. 45°
C. 25°
D. 55°
In triangle BDC,
∠B= 40, ∠D = 90
So, ∠C = 180 –(90 + 40)
= 50°
Now in triangle AEC,
∠C= 50, ∠A = 30
So, ∠E = 180 – (50 + 30)
= 100°
Thus, ∠AEB = 180 – 100 (Sum of linear pair is 180°)
= 80°
In the given figure, AB || CD, CD || EF and y : z = 3 : 7, then x = ?
A. 108°
B. 126°
C. 162°
D. 63°
Let n be the common multiple.
Y + Z = 180
3n + 7n = 180
N =18
So, y = 3n = 54o
z = 7n = 126o
x = z (Pair of alternate angles)
So, x = 126o
In the given figure, AB || CD || EF, EA ⊥ AB and BDE is the transversal such that ∠DEF = 55°. Then ∠AEB = ?
A. 35°
B. 45°
C. 25°
D. 55°
According to question
AB || CD || EF and
So, ∠D = ∠B (Corresponding angles)
According to question CD || EF and BE is the traversal then,
∠D + ∠E = 180 (Interior angle on the same side is supplementary)
So, ∠D = 180 – 55 = 125o
And ∠B = 125o
Now, AB || EF and AE is the traversal.
So, ∠BAE + ∠FEA = 180 (Interior angle on the same side of traversal is supplementary)
90 + x + 55 = 180
X + 145 = 180
X= 180 – 145 = 35o
In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?
A. 20°
B. 25°
C. 15°
D. 30°
In triangle ABC,
∠B = 70o
∠C = 20o
So, ∠A = 180o – 70o – 20o = 90o
According to question, AN is bisector of ∠A
So, ∠BAN = 45o
Now, in triangle BAM,
∠B = 70o
∠M = 90o
∠BAM = 180o – 70o – 90o = 20o
Now, ∠MAN = ∠BAN – ∠BAM
= 45o – 20o
= 25o
An exterior angle of a triangle is 110° and one of its interior opposite angles is 45°, then the other interior opposite angle is
A. 45°
B. 65°
C. 25°
D. 135°
Exterior angle formed when the side of a triangle is produced is equal to the sum of the interior opposite angles.
Exterior angle = 110°
One of the interior opposite angles = 45°
Let the other interior opposite angle = x
110° = 45° + x
x = 110° – 45°
x = 65°
Therefore, the other interior opposite angle is 65°.
The sides BC, CA and AB of ΔABC have been produced to D, E and F respectively as shown in the figure, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Then, ∠ACD + ∠BAE + ∠CBF = ?
A. 240°
B. 300°
C. 320°
D. 360°
In Δ ABC,
we have CBF = 1 + 3 ...(i) [exterior angle is equal to the sum of opposite interior angles] Similarly, ACD = 1 + 2 ...(ii)
and BAE = 2 + 3 ...(iii)
On adding Eqs. (i), (ii) and (iii),
we get CBF + ACD + BAE =2[ 1 + 2 + 3] = 2 × 180° = 4 × 90°
[by angle sum property of a triangle is 180°] CBF + ACD + BAE = 4 right angles
Thus, if the sides of a triangle are produced in order, then the sum of exterior angles so formed is equal to four right angles = 360°
The angles of a triangle are in the ratio 3:5:7. The triangle is
A. acute angled
B. right – angled
C. obtuse angled
D. isosceles
Let x be the common multiple.
3x + 5x + 7x = 180
15x = 180
x = 180/15
x = 12
3x = 3 X 12 = 36
5x = 5 X 12 = 60
7x = 7 X 12 = 84
Since, all the angles are less than 90o. So, it is acute angled triangle.
If the vertical angle of a triangle is 130°, then the angle between the bisectors of the base angles of the triangle is
A. 65°
B. 100°
C. 130°
D. 155°
Let x and y be the bisected angles.
So in the original triangle, sum of angles is
130 + 2x + 2y = 180
2(x + y) = 50
x + y= 25
In the smaller triangle consisting of the original side opposite 130 and the 2 bisectors,
x + y + Base Angle = 180
25 + Base Angle = 180
Base Angle = 155o
The sides BC, BA and CA of ΔABC have been produced to D, E and F respectively, as shown in the given figure. Then, ∠B = ?
A. 35o
B. 55o
C. 65o
D. 75o
BAC = 35o (opposite pair of angles)
BCD = 180 – 110 = 70o (linear pair of angles)
Now, in Triangle ABC we have,
A + B + C = 180o
35 + B + 70 = 180
B = 180 – 105 = 75o
In the adjoining figure, y = ?
A. 36°
B. 54°
C. 63°
D. 72°
x + y + 90 = 180 (sum of angles on a straight line)
x + y = 90 ………………….(i)
3x + 72 = 180 (sum of angles on a straight line)
3x = 108
x = 108/3 =36O
Putting this value in eq (i), we get
x + y = 90
36 + y = 90
Y = 90 – 36 = 54O
Each question consists of two statements, namely, Assertion (A) and Reason (R). Choose the correct option.
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
Sum of triangle is = 180°
And 70 + 60 + 50 = 180°
Each question consists of two statements, namely, Assertion (A) and Reason (R). Choose the correct option.
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
According to linear pair of angle, sum of angles on straight line is 180
And 90 + 90 = 180o
Each question consists of two statements, namely, Assertion (A) and Reason (R). Choose the correct option.
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
No, this is not linked with the given reason.
Each question consists of two statements, namely, Assertion (A) and Reason (R). Choose the correct option.
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
Because when two lines intersect each other, then vertically opposite angles are always equal.
Each question consists of two statements, namely, Assertion (A) and Reason (R). Choose the correct option.
A. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
B. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
C. Assertion (A) is true and Reason (R) is false.
D. Assertion (A) is false and Reason (R) is true.
3 and 5 are pair of consecutive interior angles. It is not necessary to be always equal.
Match the following columns:
The correct answer is:
A. – …….., B. – ……….,
C. – ………, D. – ……….,
(a) – (r), (b) – (s), (c) – (p), (d) – (q)
(a) – (r)
X + y = 90
X + 2x/3 = 90
5x/3 = 90
X = 270/5
= 54
(b) – (s)
X + y = 180 (according to question x =y)
X + x = 180
2x = 180
X = 90
(c) – (p)
X + y = 90 (according to question x =y)
X + x = 90
2x = 90
X = 45
(d) – (q)
X + y = 180 (linear pair of angles) ……………………….(i)
X – y =60 (according to question) …………………….. (ii)
Adding (i) and (ii) we get,
2x = 240
X = 120
Now putting this in (ii) we get,
Y = 120 – 60 = 60
Match the following columns:
The correct answer is:
A. – …….., B. – ……….,
C. – ………, D. – ……….,
(a) – (r), (b) – (p), (c) – (s), (d) – (q)
(a) – (r)
2x + 3x = 180 (linear pair of angles)
5x =180
X = 36
2x = 2 X 36 = 72
(b) – (p)
2x – 10 + 3x – 10 = 180 (linear pair of angles)
5x – 20 =180
5x = 200
x = 40
AOD = 3x – 10 (opposite angles are equal)
= 120 – 10
= 110
(c) – (s)
C = 180 – (A + B) (sum of angles triangle is 180)
= 180 – (60 + 65)
= 55
ACD = 180 – 55 (sum of linear pair of angles is 180)
= 180 – 55
= 125
(d) – (q)
B = D) (alternate interior angles)
= 55
ACB = 180 – (55 + 40) (sum of angles of triangle is 180)
= 180 – 95
= 85
The angles of a triangle are in the ratio 3:2:7. Find the measure of each of its angles.
Let x be the common multiple.
3x + 2x + 7x = 180
12x = 180
X = 15
3x = 45o
2x = 30o
7x = 105o
In a ΔABC, if ∠A – ∠B = 40° and ∠B – ∠C = 10°, find the measure of ∠A, ∠B and ∠C.
A = B + 40
C = B – 10
A + B + C = 180
B + 40 + B + B – 10 = 180
3B + 30 = 180
3B = 180 – 30 = 150
B = 50O
So, A = B + 40 = 90O
C = B – 10 = 40O
The side BC of ΔABC has been increased on both sides as shown. If ∠ABD = 105° and ∠ACE = 110°, then find ∠A.
B = 180 – 105 (sum of linear pair of angles is 180)
= 75
C = 180 – 110 (sum of linear pair of angles is 180)
= 70
So, A = 180 – (B + C) (sum of angles of triangle is 180)
= 180 – (70 + 75)
= 35O
Prove that the bisectors of two adjacent supplementary angles include a right angle.
In Δ ABC,
If one angle of a triangle is equal to the sum of the two other angles, show that the triangle is right – angled.
Let ∠𝐴 = x, ∠B = y and ∠C = z
∠𝐴 + ∠B + ∠C = 180 (sum of angles of triangle is 180)
x + y + z = 180 ……………i)
According to question,
x = y + z ………….(ii)
Adding eq (i) and (ii), we get
x + x = 180
2x = 180
X = 90
Hence, It is a right angled triangle.
In the given figure, ACB is a straight line and CD is a line segment such that ∠ACD = (3x – 5)° and ∠BCD = (2x + 10)°. Then, x = ?
A. 25
B. 30
C. 35
D. 40
3x – 5 + 2x + 10 = 180 (linear pair of angles)
5x + 5 =180
5x = 175
X = 175/ 5 = 35
In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3xׄ°, then x = ?
A. 20
B. 25
C. 30
D. 35
40 + 4x + 3x = 180 (sum of angles on a straight line)
7x + 40 =180
7x = 180 – 40
X = 140/ 7 = 20
The supplement of an angle is six times its complement. The measure of this angle is
A. 36°
B. 54°
C. 60°
D. 72°
Let x be the angle then,
complement = 90 – x
supplement = 180 – x
According to question,
180 – x = 6(90 – x)
180 – x = 540 – 6x
180 + 5x = 540
5x = 360
x = 72O
In the given figure, AB || CD || EF. If ∠ABC = 85°, ∠BCE = x° and ∠CEF = 130°, then x = ?
A. 30
B. 25
C. 35
D. 15
According to question,
AB || EF
EF || CD (AB is produced to F, CF is traversal)
FEC=130o
Now, BFC + BFO = 180o(Sum of angles of Linear pair is 180o)
BFO = 180o – 130o = 50o
Now in triangle BOF, we have
ABO = BFO + BOF
85 = 50 + BOF
BOF = 85 – 50 =35o
So, x = o
In the given figure, AB || CD, ∠BAD = 30° and ∠ECD = 50°. Find ∠CED.
A = D (Pair of alternate angles)
= 30O
Now, in triangle EDC we have
D = 30O and C = 50O
So,
CED = 180 – (C + D)
= 180 – 30 – 50
=100O
In the given figure, BAD || EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.
According to question EF || BAD
Producing E to O, we get
EFA + AEO = 180 (Linear pair of angles)
AEO = 180 – 55
= 125
Now, in triangle ABC we get,
A = 125 and C = 25
So, ABC = 180 – (A + C)
= 180 – (125 + 25)
= 180 – 150
= 30O
In the given figure, BE ⊥ AC, ∠DAC = 30° and ∠DBE = 40°. Find ∠ACB and ∠ADB.
In triangle BEC we have,
B = 40O and E = 90O
So,C = 180O – (90 + 40)
=50O
Therefore,ACB = 50O
Now intriangle ADC we have,
A = 30O and C = 50O
So,D = 180O – (30 + 50)
=100O
Therefore,
ADB + ADC = 180 (sum of angles on straight line)
ADB + 100 = 180
ADB = 180 – 100
= 80O
In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35° and QP ⊥ EF, find the measure of ∠PQH.
EGB = QHP (Alternate Exterior Angles) = 35O
QPH = 90O
So, in triangle QHP we have,
QPH + QHP + PQH = 180O
90O + 35O + PQH = 180O
PQH = 180O – 90O – 35O
= 55O
In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130°, find ∠EGF.
GEC = 180 – 130 = 50O (linear pair of angles)
According to question,
AB || CD and EF is perpendicular to AB.
GEC = EGF (pair of alternate interior angles)
= 50O
Match the following columns:
The correct answer is:
A. – …….., B. – ……….,
C. – ………, D. – ……….,
(a) – (q), (b) – (r), (c) – (s), (d) – (p)
(a) – (q)
x + x + 10 = 90
2x + 10 = 90
2x = 80
x = 40
x + 10 = 50O
(b) – (r)
A + B + C =180
65 + B + B – 25 = 180
2B + 40 = 180
2B = 140
B = 70O
(d) – (p)
A + B + C + D =360
2x + 3x + 5x + 40 = 360
10x + 40 = 360
10x = 320
X = 32O
5x = 32 X 5 = 160O
In the given figure, lines AB and CD intersect at O such that ∠AOD + ∠BOD + ∠BOC = 300°. Find ∠AOD.
According to question,
In the given figure CD is a straight line.
As we know, Sum of angle on a straight line is 180O
S0,
AOD + BOD + BOC =300
AOD + 180 =300
AOD =300 – 180
= 120O
In the given figure AB || CD, ∠APQ = 50° and ∠PRD = 120°. Find ∠QPR.
According to question,
PRD = 120O
PRD = APR (Pair of alternate interior angles)
So,
APR = 120
APQ + QPR = 120
50 + QPR = 120
QPR = 120 – 50
= 70O
In the given figure, BE is the bisector of ∠B and CE is the bisector of ∠ACD.
Prove that
In triangle ABC we have,
A + B + C = 180
Let B = x and C = y then,
A + 2x + 2y = 180 (BE and CE are the bisector of angles B and C respectively.)
x + y + A = 180
A = 180 – (x + y) ………….(i)
Now, in triangle BEC we have,
B = x/2
C = y + ((180 – y) / 2)
= (180 + y) / 2
B + C + BEC = 180
x/2 + (180 + y) / 2 + BEC = 180
BEC = (180 – x – y) /2 ………..(ii)
From eq (i) and (ii) we get,
BEC = A/2
In ΔABC, sides AB and AC are produced to D and E respectively. BO and CO are the bisectors of ∠CBD and ∠BCE respectively. Then, prove that
Here BO, CO are the angle bisectors of ∠DBC &∠ECB intersect each other at O.
∴∠1 = ∠2 and ∠3 = ∠4
Side AB and AC of ΔABC are produced to D and E respectively.
∴ Exterior of ∠DBC = ∠A + ∠C ………… (1)
And Exterior of ∠ECB = ∠A + ∠B ………… (2)
Adding (1) and (2) we get
∠DBC + ∠ECB = 2 ∠A + ∠B + ∠C.
2∠2 + 2∠3 = ∠A + 180°
∠2 + ∠3 = (1 /2)∠A + 90° ………… (3)
But in a ΔBOC = ∠2 + ∠3 + ∠BOC = 180° ………… ( 4)
From eq (3) and (4) we get
(1 /2)∠A + 90° + ∠BOC = 180°
∠BOC = 90° – (1 /2)∠A
Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.
Let x be the common multiple.
So, angles will be x, 2x and 3x
X + 2x + 3x = 180
6x = 180
X =30
2x = 2 X 30 = 60
3x = 3 X 30 = 90
So, Angles are 30O,60O and 90O
In ΔABC, ∠B = 90° and BD ⊥ AC. Prove that ∠ABD = ∠ACB.
Let ∠ABD = x and ∠ACB = y
According to question,
∠B = 90O
In triangle BDC, we have,
∠BDC = 90O
∠DBC = (90 – x)O
∠BDC + ∠DBC + ∠DCB = 180O
90O + (90 – x)O + y = 180O
180O – x + y = 180O
x = y
So,
∠ABD = ∠ACB