T-ratios Of Some Particular Angles
Class 10th Mathematics RS Aggarwal Solution
- sin 60 cos 30 + cos 60 sin 30 Evaluate each of the following:
- cos 60 cos 30 sin 60 sin 30 Evaluate each of the following:
- cos 45 cos 30 + sin 45 sin 30 Evaluate each of the following:
- sin30^circle /cos45^circle + cot45^circle /sec60^circle - sin60^circle…
- 5cos^260^circle + 4sec^230^circle - tan^245^circle /sin^230^circle +…
- 2 cos^2 60 + 3 sin^2 45 3 sin^2 30 + 2 cos^2 90 Evaluate each of the following:…
- cot^2 30 2cos^2 30 (3/4)sec^2 45 + (1/4)cosec^2 30 Evaluate each of the…
- (sin^2 30 + 4 cot^2 45 sec^2 60)(cosec^2 45 sec^2 30) Evaluate each of the…
- 4/cot^230^circle + 1/sin^230^circle - 2cos^245^circle - sin^20^circle Evaluate…
- Show that: (i) 1-sin60^circle /cos60^circle = tan60^circle - 1/tan60^circle + 1…
- Verify each of the following: (i) sin 60° cos 30° — cos 60° sin 30° = sin 30°…
- If A = 45°, verify that: (i) sin 2A = 2 sin A cos A (ii) cos 2A = 2 cos^2 A —1…
- If A = 30°, verify that: (i) sin2a = 2tana/1-tan^2a (ii) cos2a =
- (i) sin (A + B) = sin A cos B + cos A sin B (ii) cos (A + B) = cos A cos B —…
- (i) sin (A — B) = sin A cos B — cos A sin B (ii) cos (A — B) = cos A cos B +…
- If A and B are acute angles such that tan a = 1/3 , tanb = 1/2 tan (a+b) =…
- Using the formula, tan2a = 2tana/1-tan^2a , find the value of tan 60°, it being…
- Using the formula, cosa = root 1+cos2a/2 , find the value of cos 30°, it being…
- Using the formula, sina = root 1-cos2a/2 , find the value of sin 30°, it being…
- In the adjoining figure, ΔABC is a right-angled triangle in which B = 90°, ∠A=…
- In the adjoining figure, ΔABC ABC is right-angled at B and ∠A= 30°. If BC = 6…
- In the adjoining figure, ΔABC is right-angled at B and ∠A= 45°. If AC = 3√2 cm,…
- If sin (A + B) =1 and cos (A - B) = 1, 0° ≤ (A + B) ≤ 90° and A B then find A…
- If sin (A - B) = 1/2 and cos (A +B) = 1/2 ,0° (A + B) 90° and A B then find A…
- If tan (A - B) = 1/√3 and tan (A + B) = √3, 0° (A + B) 90° and A B then find A…
- If 3x = cosec θ and 3/x = cot θ, find the value of 3 (x^2 - 1/x^2)…
- If sin (A + B) = sin A cos B + cos A sin B , and cos (A - B) = cos A cos B +…
Exercise 6
Question 1.Evaluate each of the following:
sin 60° cos 30° + cos 60° sin 30°
Answer:
Since sin 60° = √3/2 = cos 30°
and sin 30° = 1/2 = cos 60°
∴ sin 60° cos 30° + cos 60° sin 30° = (√3/2)( √3/2) + (1/2)(1/2)
= (3/4) + (1/4)
= 4/4
= 1
Question 2.
Evaluate each of the following:
cos 60° cos 30° — sin 60° sin 30°
Answer:
Since cos 60° = 1/2 = sin 30°
and cos 30° = √3/2 = sin 60°
∴ cos 60° cos 30° — sin 60° sin 30° = (1/2) × (√3/2) - (√3/2) × (1/2)
= (√3/4) - (√3/4)
= 0
Question 3.
Evaluate each of the following:
cos 45° cos 30° + sin 45° sin 30°
Answer:
Since cos 45° = 1/√2 = sin 45°
cos 30° = √3/2
sin 30° = 1/2
∴ cos 45° cos 30° + sin 45° sin 30° = (1/√2)×(√3/2) + (1/√2)(1/2)
= (√3 / 2√2) + (1 / 2√2)
= (√3 + 1) / (2√2)
Question 4.
Evaluate each of the following: 
Answer:
Since sin 30° = 1/2 , sin 60° = √3/2 , sin 90° = 1
cos 30° = √3/2 , cos 45° = 1/√2 , cos 60° = 1/2
sec 60° = ( 1/cos 60° )= 2
tan 45° = (sin 45° /cos 45°) = (1/√2) / ( 1/√2) = 1
cot 45° = (1/tan 45°) = 1/1 = 1
∴ 
= ((1/2) / ((1/√2)) + (1 /2 ) – ((√3/2) / 1) + ((√3/2)/1)
= (1/√2) + (1/2) – (√3/2) + (√3/2)
= (1/√2) + (1/2)
= (√2/2) + (1/2)
= (√2 + 1)/2
Question 5.
Evaluate each of the following: 
Answer:
cos 30° = √3/2 ⇒ cos2 30° = 3/4
cos 60° = 1/2 ⇒ cos2 60° = 1/4
sec 30° = (1/cos 30°) = (2/√3) ⇒ sec2 30° = 4/3
tan 45° = 1 ⇒ tan2 45° = 1
sin 30° = 1/2 ⇒ sin2 30° = 1/4
We also know that sin2 θ + cos2 θ = 1
∴ 
= [(5×(1/4)) + (4×(4/3)) – (1)]/1
= (5/4) + (16/3) – 1
= (15 + 64 - 12)/12 = 67/12
Question 6.
Evaluate each of the following:
2 cos260° + 3 sin245° — 3 sin230° + 2 cos290°
Answer:
cos 60° = 1/2 ⇒ cos2 60° = 1/4
sin 45° = 1/√2 ⇒ sin2 45° = 1/2
sin 30° = 1/2 ⇒ sin2 30° = 1/4
cos 90° = 0 ⇒ cos2 90° = 0
∴ 2 cos260° + 3 sin245° — 3 sin230° + 2 cos290°
= 2(1/4) + 3(1/2) – 3(1/4) + 2(0)
= (1/2) + (3/2) – (3/4) = 2 – (3/4)
= 5/4
Question 7.
Evaluate each of the following:
cot230° — 2cos230° — (3/4)sec245° + (1/4)cosec230°
Answer:
cos 30° = √3/2 , ⇒ cos2 30° = 3/4
sin 30° = 1/2
∴ cosec 30° = 1/sin30° = 2 ⇒ cosec2 30° = 4
cot 30° = (cos 30°/sin 30°) = √3 ⇒ cot2 30° = 3
cos 45° = 1/√2
∴ sec 45° = 1/cos 45° = √2 ⇒ sec2 45° = 2
∴ cot230° — 2cos230° — (3/4)sec245° + (1/4)cosec230°
= 3 – 2(3/4) – (3/4) × 2 + (1/4) × 4
= 3 – 1.5 – 1.5 + 1
= 1
Question 8.
Evaluate each of the following:
(sin230° + 4 cot245° — sec260°)(cosec245° sec230°)
Answer:
sin 30° = 1/2 ⇒ sin2 30° = 1/4
cos 45° = 1/√2 = sin 45°
cot 45° = 1 ⇒ cot2 45° = 1
cos 60° = 1/2 ⇒ sec 60° = 2 ⇒ sec2 60° = 4
cos 30° = √3/2 ⇒ sec 30° = 2/√3 ⇒ sec2 30° = 4/3
cosec 45° = 1/sin 45° = √2 ⇒ cosec2 45° = 2
∴ (sin230° + 4 cot245° — sec260°)(cosec245° sec230°)
= ((1/4) + 4(1) – 4) × (2)(4/3)
= (1/4) × (8/3)
= 8/12
= 2/3
Question 9.
Evaluate each of the following: 
Answer:
sin 30° = 1/2 , ⇒ sin2 30° = 1/4 ⇒ (1/sin2 30°) = 4
cos 30° = √3/2 ,
cot 30° = (cos 30°/ sin 30°) = √3 ⇒ cot2 30° = 3
cos 45° = 1/√2 ⇒ cos2 45° = 1/2
sin 0° = 0
- 2 cos2 45° - sin2 0° = (4/3) + (4) – 2(1/2) – 0
= (4/3) + 4 – 1
= 13/3
Question 10.
Show that:
(i) 
(ii) 
Answer:
(i) Consider L.H.S. = 
= 
= (2-√3)
Consider R.H.S. = 
= 
= 
(Rationalizing the denominator)
= 
= 2 - √3
L.H.S. = R.H.S.
Hence, verified.
(ii) L.H.S. = 
R.H.S. = cos 30° = √3/2
L.H.S. = R.H.S.
Hence, verified.
Question 11.
Verify each of the following:
(i) sin 60° cos 30° — cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°
Answer:
(i) Consider L.H.S. = sin 60° cos 30° — cos 60° sin 30°
= (√3/2) × (√3/2) – (1/2)(1/2)
= (3/4) – (1/4)
= 2/4
=1/2
Consider R.H.S. = sin 30° = 1/2
L.H.S. = R.H.S.
Hence, verified.
(ii) Consider L.H.S. = cos 60° cos 30° + sin 60° sin 30°
= (1/2) × (√3/2) + (√3/2)(1/2)
= (√3/4) + (√3/4)
= √3/2 = cos 30° = R.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
(iii) Consider L.H.S. = 2 sin 30° cos 30°
= 2 × (1/2) × (√3/2)
= √3/2 = sin 60° = R.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
(iv) Consider L.H.S. = 2 sin 45° cos 45°
= 2 × (1/√2) × (1/√2)
= (2 × 1/2)
= 1 = sin 90° = R.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
Question 12.
If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = 2 cos2A —1 = 1— 2 sin2A
Answer:
(i) To show: sin 2A = 2 sin A cos A
A = 45°
∴ To show: sin 90° = 2 sin 45° cos 45°
Consider R.H.S. = 2 sin 45° cos 45°
= 2 × (1/√2) × (1/√2)
= (2 × 1/2)
= 1 = sin 90° = L.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
(ii) To show: cos 2A = 2 cos2A —1 = 1— 2 sin2A
A = 45°
∴ To show: cos 90° = 2 cos2 45° —1 = 1— 2 sin2 45°
Consider 2 cos2 45° —1 = 1— 2 sin2 45°
= 2 × (1/√2) – 1
= 1 – 1
= 0
= cos 90° = R.H.S.
Consider 1— 2 sin2 45° = 1 – 2(1/2)
= 1 – 1
= 0
= cos 90° = R.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
Question 13.
If A = 30°, verify that:
(i) 
(ii) 
(iii) 
Answer:
(i) To prove:- sin 2A = 
A = 30°
∴ To show:- sin 60° = 
Consider L.H.S. = 
= 
= 
= 
= 
= √3/2
= sin 60° = R.H.S.
∴ R.H.S. = L.H.S.
Hence, verified.
(ii) To show:- cos 2A = 
A = 30°
∴ To show:- cos 60° = 
Consider R.H.S. = = 
= 2/4 = 1/2 = cos 60° = L.H.S.
∴ R.H.S. = L.H.S.
Hence, verified.
(iii) To show:- tan2A =
A = 30°
∴ To show:- tan 60° = 
Consider R.H.S. = 
= 
= 
= 3/√3 = √3
= tan 60° = L.H.S.
∴ R.H.S. = L.H.S.
Hence, verified.
Question 14.
If A = 60° and B = 30°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B — sin A sin B
Answer:
(i) To verify: sin (A + B) = sin A cos B + cos A sin B
If A = 60° and B = 30°, then
To verify: sin 90° = sin 60° cos 30° + cos 60° sin 30°
Consider R.H.S. = sin 60° cos 30° + cos 60° sin 30°
= (√3/2) × (√3/2) + (1/2)(1/2)
= (3/4) + (1/4)
= 4/4
=1
Consider L.H.S. = sin 90° = 1
∴ L.H.S. = R.H.S.
Hence, verified.
(ii) To verify: cos (A + B) = cos A cos B — sin A sin B
If A = 60° and B = 30°, then
To verify: cos (90°) = cos 60° cos 30° — sin60° sin 30°
Consider R.H.S. = cos 60° cos 30° - sin 60° sin 30°
= (1/2) × (√3/2) - (√3/2)(1/2)
= (√3/4) - (√3/4)
= 0 = cos 90° = L.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
Question 15.
If A = 60° and B = 30°, verify that:
(i) sin (A — B) = sin A cos B — cos A sin B
(ii) cos (A — B) = cos A cos B + sin A sin B
(iii) 
Answer:
(i)To verify: sin (A - B) = sin A cos B - cos A sin B
If A = 60° and B = 30°, then
Consider LHS
sin(60°-30°) = sin 30° = 1/2
Consider R.H.S. = sin 60° cos 30° - cos 60° sin 30°
= (√3/2) × (√3/2) - (1/2) (1/2)
= (3/4) - (1/4)
= 2/4
=1/2
∴ L.H.S. = R.H.S.
Hence, verified.
(ii) To verify: cos (A - B) = cos A cos B + sin A sin B
If A = 60° and B = 30°, then
To verify: cos (30°) = cos 60° cos 30° + sin 60° sin 30°
Consider R.H.S. = cos 60° cos 30° + sin 60° sin 30°
= (1/2) × (√3/2) + (√3/2)(1/2)
= (√3/4) + (√3/4)
= √3/2 = cos 30° = L.H.S.
∴ L.H.S. = R.H.S.
Hence, verified.
(iii) To verify:- tan(A +B) =
If A = 60° and B = 30°, then
To verify: tan 90° = 
Consider R.H.S. = 
= 
= 
= 4/0 = ∞
= tan 90° = L.H.S.
∴ R.H.S. = L.H.S.
Hence, verified.
Question 16.
If A and B are acute angles such that tan 
, show that A + B = 45°.
Answer:
Given: tan (A +B) = and tan A = 1/3, tan B = 1/2
∴ tan(A +B) =
= 
= 
= 1
∴ tan(A + B) = 1
Also, A and B are acute angles, therefore both A and B are less than 90°. So A + B must be less than 180°.
Therefore, the only possible case for which tan (A+B) = 1 will be when (A + B) equals 45°.
Thus, A + B = 45°
Question 17.
Using the formula, 
, find the value of tan 60°, it being given that 
Answer:
To find:- tan 60°
Given: tan2A = ……………………(1)
tan 30° = 1/√3
∴ Putting A = 30° in equation (1), we have the following:
tan 60° =
∴ tan 60° = = =
= 3/√3 = √3
∴ tan 60° = √3
Question 18.
Using the formula,
, find the value of cos 30°, it being given that cos 60° = 1/2 .
Answer:
Given: cos A = 
, …………………….(1)
cos 60° = 1/2
To find: cos 30°
By putting A = 30° in equation (1), we get the following:
cos 30° = 
= 
= 
= √3/2
∴ cos 30° = √3/2
Question 19.
Using the formula, 
, find the value of sin 30°, it being given that cos 60° = 
Answer:
Given: sin A = 
, …………………….(1)
cos 60° = 1/2
To find: sin 30°
By putting A = 30° in equation (1), we get the following:
sin 30° = 
= 
= 
= 1/2
∴ sin 30° = 1/2
Question 20.
In the adjoining figure, ΔABC is a right-angled triangle in which B = 90°, ∠A= 30° and AC = 20 cm. Find (i) BC, (ii) AB.
Answer:
Since, in a right angled triangle,
sin θ = Perpendicular / Hypotenuse ,
and cos θ = Base / Hypotenuse ,
where θ is the angle made between the hypotenuse and the base.
(i) ∴ In the given figure, sin 30° = BC/AC
⇒ 1/2 = BC/20
⇒ (1/2) × 20 = BC
⇒ BC = 10 cm
(ii) Now, In the given figure, cos 30° = AB/AC
⇒ √3/2 = AB/20
⇒ (√3/2) × 20 = AB
⇒ AB = 10√3 cm
Question 21.
In the adjoining figure, ΔABC ABC is right-angled at B and ∠A= 30°. If BC = 6 cm, find (i) AB, (ii) AC.
Answer:
Since, in a right-angled triangle,
sin θ = Perpendicular / Hypotenuse,
and cos θ = Base / Hypotenuse,
where θ is the angle made between the hypotenuse and the base.
(i) ∴ In the given figure, sin 30° = BC/AC
⇒ 1/2 = 6/AC
⇒ AC = 6 × 2
⇒ AC = 12 cm
(ii) Now, In the given figure, cos 30° = AB/AC
⇒ √3/2 = AB/12
⇒ (√3/2) × 12 = AB
⇒ AB = 6√3 cm
Aliter: Since ABC is a right-angled triangle,
∴ (AB)2 +(BC)2 = (AC)2
∴ (AB)2 = (AC)2 - (BC)2
⇒ (AB)2 = 144 – 36 = 108
⇒ (AB) = √108 = 6√3
∴ AB = 6√3 cm
Question 22.
In the adjoining figure, ΔABC is right-angled at B and ∠A= 45°. If AC = 3√2 cm, find (i) BC, (ii) AB.
Answer:
Since, in a right-angled triangle,
sin θ = Perpendicular / Hypotenuse,
and cos θ = Base / Hypotenuse,
where θ is the angle made between the hypotenuse and the base.
(i) ∴ In the given figure, sin 45° = BC/AC
⇒ 1/√2 = BC / (3√2)
⇒ BC = (1/√2) × (3√2) = 3
⇒ BC = 3 cm
(ii) Now, In the given figure, cos 45° = AB/AC
⇒ 1/√2 = AB / (3√2)
⇒ AB = 1/√2 × (3√2)
⇒ AB = 3 cm
Question 23.
If sin (A + B) =1 and cos (A - B) = 1, 0° ≤ (A + B) ≤ 90° and A > B then find A and B.
Answer:
Given: (i) sin (A + B) = 1
(ii) cos (A - B) = 1
Since, sin (A + B) = 1
⇒ sin (A + B) = sin 90° (∵ 0° ≤ (A + B) ≤ 90°, sin 90° = 1)
⇒ A + B = 90° .......................................... (1)
Also, cos (A - B) = 1
⇒ cos (A - B) = cos 0° (∵ 0° ≤ (A + B) ≤ 90°, cos 0° = 1)
⇒ A - B = 0° .......................................... (2)
From equation (2), we get A = B
Putting this value in equation (1), we get, 2A = 90° ⇒ A = 45°
∴ B = A = 45°
∴ 
A = B = 45°
Question 24.
If sin (A - B) = 1/2 and cos (A +B) = 1/2 ,0° < (A + B) < 90° and A > B then find A and B.
Answer:
Given: (i) sin (A - B) = 1/2
(ii) cos (A + B) = 1/2
Since, sin (A - B) = 1
⇒ sin (A - B) = sin 30° (∵ 0° < (A + B) < 90°, sin 30° = 1/2)
⇒ A - B = 30° .......................................... (1)
Also, cos (A + B) = 1/2
⇒ cos (A + B) = cos 60° (∵ 0° < (A + B) < 90°, cos 60° = 1/2)
⇒ A + B = 60° .......................................... (2)
On adding equation (1) and (2), we get,
2A = 90° ⇒ A = 45°
Putting this value in equation (2), we get,
B = 60° - A = 60° - 45° ⇒ B = 15°
∴ ∠A = 45°, ∠B = 15°
Question 25.
If tan (A - B) = 1/√3 and tan (A + B) = √3, 0° < (A + B) < 90° and A > B then find A and B.
Answer:
Given: (i) tan (A - B) = 1/√3
(ii) tan (A + B) = √3
Since, tan (A - B) = 1/√3
⇒ tan (A - B) = tan 30°
(∵ 0° < (A + B) < 90°, tan 30° = 1/√3)
⇒ A - B = 30° .......................................... (1)
Also, tan (A + B) = √3
⇒ tan (A + B) = tan 60°
(∵ 0° < (A + B) < 90°, tan 60° = √3)
⇒ A + B = 60° .......................................... (2)
On adding equation (1) and (2), we get,
A - B + A + B = 30° + 60°2A = 90°
⇒ A = 45°
Putting this value in equation (2), we get,
B = 60° - A
= 60° - 45°
⇒ B = 15°
∴ ∠A = 45°, ∠B = 15°
Question 26.
If 3x = cosec θ and 3/x = cot θ, find the value of 
Answer:
Given,
3x = cot θ
⇒ 9x2 = cot2θ [1]
and
Subtracting [2] from [1], we get
Question 27.
If sin (A + B) = sin A cos B + cos A sin B , and
cos (A - B) = cos A cos B + sin A sin B, find the values of
(i) sin 75° and (ii) cos 15°.
Answer:
Given: sin (A + B) = sin A cos B + cos A sin B
cos (A - B) = cos A cos B + sin A sin B
(i) To find: sin 75°
If we put A = 30° and B = 45°, then we have:
sin 75° = sin 30° cos 45° + cos 30° sin 45°
∴ sin 75° = (1/2) × (1/√2) + (√3/2) × (1/√2)
= (1/2√2) + (√3/2√2)
= 
(ii) To find: cos 715°
If we put A = 45° and B = 30°, then we have:
cos 15° = cos 45° cos 30° + sin 45° sin 30°
∴ cos 15° = (1/√2) × (√3/2) + (1/√2) × (1/2)
= (√3 / 2√2) + (1/2√2)
=
∴ (i) sin 75° =
(ii) cos 15° =