Some Application Of Trigonometry

Let the height of the tower = h (m)

In ∆ABC,

tan60° =

√3 =

h = 20√3 m

Therefore height of the tower is 20√3 m

Let the length of the ladder=l (m)

In ∆ABC,

cos60° =

=

l =9.5×2 ⇒ 19 m

Therefore length of Ladder is 19 m.

Let the length of the wall = h (m)

In ∆ABC,

tan60°=

√3 =

h = 2√3 m

Therefore length of the wall is 2√3 m

Let the length of the wire = l (m)

In ∆ABC,

Sin 45° =

=

l ⇒ 10√2 m

⇒ 10x1.41

⇒14.1 m

Therefore Length of wire is 14.1 m

Let the length of the wire = l (m)

In ∆ABC,

Sin 60° =

=

= 2×75

l =

l = ⇒ ⇒ 50√3 ⇒ 86.6 m.

Therefore length of string is 86.6 m.

Given: The length of a string between a kite and a point on the ground is 90 metres. If the string makes an angle with the ground level such that tan=15/8.

To find: how high is the kite.

Solution:


Let the height of string = h (m)

tan θ = (given)


Since tanθ = perpendicular/base

So perpendicular= 15 and base=8

So we construct a right triangle ABC right angled at C such that

∠ABC=θ and AC = Perpendicular = 15

BC = base = 8

By Pythagoras theorem,

AB² = AC² + BC²

⇒ AB² = (15)² + (8)²

⇒ AB² = 225 + 64

⇒ AB² = 289

⇒ AB = √289

⇒ AB = 17

⇒ Sin θ = ..... (1)

In ∆ABC,

Sin θ =

Sin θ = ...... (2)

=


17h = 90×15

⇒ h =


⇒h = 79.41 m.

Therefore length of string is 79.41 m.

Let the height of tower = h (m)

Let the height of the flag-staff = t (m)

In ∆DBC,

tan 45° =

1 =

h = 70 m

Therefore height of tower = 70m.

Now in ∆ABC,

tan 60° =

√3 = ⇒ √3 = (on substituting value of h =70)

70+t = 70√3

t = 70√3-70

t = 70 (√3 -1)

t = 70 × (1.732-1)

t = 70 × 0.732 ⇒ 51.24 m.

Therefore height of the flag- staff is 51.24 m.

Total height of the tree is 15 m.

I.e AB= 15m.

Let height at which tree is broken is h (m)

Therefore BC = h (m)

CD = AB-BC

= 15-h.

In ∆DBC,

sin 60° =

=

On cross-multiplication

√3(15-h) = 2h

⇒ 15√3-√3h = 2h

⇒ h(2+√3) = 15√3

h =

on multiplying and dividing by 2-√3

h =

h = ⇒ ⇒ 6.96 m.

Therefore the tree is broken at 6.96 m from the ground.

Let the height of the tower = h (m)

Let the point of elevation on the ground is (m) away from the foot of the tower.

In ∆DBC,

tan 30° =

=

On the cross multiplication

= h√3 -------(1)

In ∆ABC,

tan 60° =

√3 =

√3 = ---------(2)

On substituting value of from equn. (1) in eqn. (2)

√3 =

h√3×√3 = 5+h

3h = 5+h

3h-h = 5

2h = 5 ⇒ h =

h = 2.5 m.

Therefore height of the tower is 2.5 m.

Let the height of the tower = h (m)

In ∆ABD,

tan 60° =

√3 =

h = ---------- (1)

In ∆ABC,

tan 30° =

tan 30° =

------(2)

on substituting value of h from equn. (1) In equn. (2)

On cross multiplication

√3×√3 =

3 =

3

2

⇒ 25 m.

Now substituting value of in eqn. (1)

h = 25√3 ⇒ 43.3m.

Therefore height of tower is 43.3 m.

Let the height of the tower = h (m)

Let the point of 60° elevation is (m) away from the foot of the tower.

In ∆ABC,

tan 45° =

1 =

1 =

h = 10+ ----(1)

In ∆ABD,

tan 60° =

√3 =

h = √3

= ---------(2)

From eqn. (2) in eqn. (1)

h =

h - = 10

= 10

= 10√3

⇒

h= ⇒ h=

⇒ 15+5√3 ⇒ 23.66 m.

Therefore height of the tower is 23.66 m.

Let the height of the parachutist = h (m)

Let the distance of falling point from observation point = (m)

In ∆ABC,

tan 45° =

1 = ⇒

h = 100+ ----------(1)

In ∆ABD,

tan 60° = ⇒

√3 =

h = √3 -----------(2)

From eqn. (1) and eqn. (2) we get,

100+ = √3

100 = (√3-1)

= ⇒ ⇒

136.6 m.

in eqn. (1)

100+

h = 100+136.6

h = 236.6 m.

Therefore height of parachutist is 236.6 m. and distance of point where he falls

is 136.6 m.

Let the distance between the objects = (m.)

In ∆ABC,

tan 45° =

1 = ⇒

1 =

= 150 ----------(1)

In ∆ABD,

tan 60° =

√3 =

= 150

= -------(2)

substituting value of y in eqn.(1)

=

= ⇒ ⇒

=

= 63.4 m.

Therefore the distance between the points is 63.4 m.

Let the height of the tower = h (m)

In ∆ABC,

tan 30° =

=

=

√3h = 150+

-------- (1)

In ∆ABD,

tan 60° =

√3 = ⇒ h = √3

-------(2)

on substituting value of x from eqn.(2)i eqn.(1)

h-3h = -150

2h = 150√3

h = ⇒ 75

h= 129.9 m.

Hence height of tower is 129.9 m.

Let the height of the tower = h (m)

Let the distance of point from the foot of the tower = (m)

In ∆ABC,

tan 32° =

tan 32° =

0.6248 =

h = 0.6248(100+) -----------(1)

In ∆ABD,

tan 63° =

1.9626 =

h = 1.9626 -----------(2)

Substituting value of h from eqn. (2) in eqn. (1)

1.9626 = 0.6248(100+

1.9626 = 62.48+0.6248

1.9626 -0.6248 = 62.48

1.3378 = 62.48

= 46.70 m.

on substituting value of x in eqn.(2)

1.9626× 46.7

91.66 m.

Distance of the first position from tower=

= CD+DB

100+46.7

height of tower is 91.66 m.

Let the height of the tower is = h (m)

Distance of point B from foot of the tower is = (m)

In ∆ADC,

tan 30° =

=

√3 h = 20+ -----------(1)

In ∆DCB,

tan 60° =

√3 =

h = √3 ----------(2)

On substituting value of h from eqn. (2) in eqn. (1)

√3× √3 = 20 +

320+

3-= 20

= 10

Therefore distance of point A from tower is

AC = AB+BC

AC = 20+10 ⇒ 30

Ac = 30 m.

Now substituting value of in eqn. (1)

h = 20+10 ⇒ 30

⇒ 17.32 m.

Therefore height of tower is 17.32 m.

Let the distance between tower and building = (m)

In ∆ABC,

tan 60° =

√3 =

=

√3 = h+15

h = √3-15 ------(1)

In ∆ABE,

tan 30° =

---------(2)

h = 3h-15 ⇒ 2h = 15

h = ⇒ 7.5 m.

Height of tower = 15+7.5 ⇒ 22.5 m.

⇒ 12.99 m.

Therefore distance between tower and building is 12.99 m.

Let the height of the Flag-pole = h(m)

And height of tower = (m)

In ∆ABC,

tan 60° =

√3 =

h+ = 9√3 -------(1)

In ∆DBC,

tan 30° =

=

√3 = 9

⇒

-------(2)

Now substituting value of in eqn. (1)

h+3√3 = 9√3

h = 9√3-3√3

h = 6√3 m.

Therefore height of tower is 3√3 m. and height of flag pole is 6√3 m.

let the broken part be DB.

BC = 8 cm

In ∆DBC,

tan 30° =

=

---------(1)

Cos 30° =

=

h =

Height of tree =
=

=
=

Height of tree = 8√3 m

Let the height of the flag-staff = h (m)

And the distance of point P from foot of building = (m)

In ∆APB,

tan 45° =

tan 45° =

1 = =

h+10 = -------(1)

In ∆DPB,

tan 30° =

=

⇒17.32 m. -----------(2)

On substituting value of in eqn. (1)

h =

h = 17.32-10 ⇒ 7.32 m.

Therefore height of flag-staff is 7.32 m. and distance of point P from tower is 17.32 m.

Let the height of the lamp post = h (m)

And height of girl is CD = 1.6 m.

Length of shadow is OD = 4.8 m.

In ∆CDO

= ⇒ ⇒ ------(1)

In ∆ABO

=

=

By Smilariting

In ∆ABO and in ∆CDO

∠ABO = ∠CDO = 90°

∠AOB = ∠COD

∆ABO ∆CDO (AA similarity)

For similar ∆s sides are in ratio

Hance,

=

AP = AB - BP

= 30-1.5
= 28.5 m.

In ∆ADP,

DP = 28.5√3 ------------(1)

_{In ∆AEB,}

= 19√3 m.

Therefore the walking distance of boy is 19√3 m.

Let the height of tower = h (m)
Since the tower is vertical to the ground.
∠ABC = 90°
We know, in a right-angle triangle,

_{In ∆ABC,}

=

=

√3h = 40 ---------------(1)

_{In ∆ABD,}

√3 =

h =√3 -------(2)

on substituting the value of h from eqn. (2) in eqn. (1)

h = √3× (√3h-40)

h = 3h-40√3

2h = 40√3

h = 20√3 m.

Therefore the height of the tower is 20√3 m.

Since the building is vertical.
∠QPO = 90°

_{In ∆OPQ,}

tan 45° =

1 =

OP = 20 -------(1)

Now in ∆OPR

tan 60° =

=

=

20√3 = h+20

h = 20√3-20

h = 20(√3-1) m.

Therefore the height of transmission tower is 20(√3-1) m.

Let DC is tall building and AB is multistoried building.

AB = AE+EB

AB = h+8 ----------(1)

_{In ∆ABC,}

tan 45° =

1 =

----------(2)

_{In ∆AED,}

tan 30° =

=

√3h = ---------(3)

Substituting value of from eqn. (2) in eqn. (1)

√3h-h = 8

h =

h = ⇒

h = 4(√3+1)m. --------(4)

Substituting value of h from eqn. (4) in eqn. (3)

√3h =

√3× 4(√3+1)

√3 (4√3+4)

Therefore height of multistoried building is

= 8+4(√3+1)

= 8+4√3+4

= 12+4√3

= 4(3+√3) m.

Distance between two building is 4(3+√3) m.

Let AD is statue of height 1.6 m. and BD is pedestal of height h (m).

Let the distance between point of elevation and foot of pedestal is (m).

_{In ∆DBC,}

tan 45° =

1 =

h = ----------(1)

_{In ∆ABC,}

tan 60° =

√3 =

√3 =

√3 = 1.6+h -----------(2)

On substituting value of from eqn. (1) in eqn. (2)

√3h = 1.6+h

√3h – h = 1.6

h =

on rationalizing we get.

h = ⇒ ⇒

= = m.

Therefore height of pedestal is m.

Let BC be the height of the T.V tower and AB be the width of the river.

_{In ∆ABC,}

tan 60° =

√3 = ------(1)

_{In ∆ABC,}

tan 30° =

tan 30° =

=

√3h = 20+ ---------(2)

On substituting value of h in eqn.(2)

√3× √3 = 20+

3

3

2

On substituting value of in eqn. (1)

h = √3

h = 10√3 m.

Therefore height of T.V tower is 10√3 m. and width of river is 10 m.

Let the distance between the foots of building and cable tower is (m).

The height of cable tower = AB = AE+EB ⇒ (h+7)m.

_{In ∆AED,}

tan 60° =

√3 =

h = √3 ---------(1)

The height of cable tower = AB = AE+EB ⇒ (h+7)m.

_{In ∆DEB,}

tan 45° =

1 =

----------(2)

On substituting value of in eqn. (1)

h = 7√3

Height of cable tower is (h+7)m.

⇒ 7√3+7

⇒ 7(√3+1)m.

Therefore height of cable tower is 7(√3+1)m.

Let the distance between the two ships be (m.)

And distance between the ship and foot of light house is (m.)

_{In ∆ABD}

tan 45° =

1 =

AB = BD

75 =

--------(1)

_{In ∆ABC}

tan 30° =

tan 30° =

= 75√3 --------(2)

On substituting value of in eqn. (2)

Therefore distance between the two ships is

Let AB be the building of height 50 m. and tower of height h (m.)

_{In ∆ABC}

tan 60° =

√3 =

------(1)

_{Now in ∆DCB}

tan 30° =

-----(2)

On substituting value of in eqn. (1)

√3h =

h =

h =

Therefore height of tower is m.

Let A and B be the points on the bank n opposite sides f the river and BC be the width of the river.

BC = BD+DC ⇒ (m.

AD be the height of the bridge.

In ∆ADC

tan 45° =

1

---------(1)

In ∆ADB

tan 30° =

------------(2)

Width of the river = (

⇒ 30√3+30

⇒ 30(√3+1)m.

Let AB and ED are two poles of equal height.

Let C be the point between the poles on the ground.

In ∆EDC

tan 60° =

h = √3 --------(1)

In ∆ABC

tan 30° =

√3h = 80-x----------(2)

On substituting value of h from eqn.(1) in eqn. (2)

√3 × √3x = 80 - x

⇒ 4x = 80

On substituting value of in eqn. (1)

h = 20√3
Distance of C from pole ED = 20 m
Distance of C from pole AB = 80 - 20
= 60 m

Therefore the height of the poles is 20√3 m. and distances of the points from one pole is 20 m and from other pole is 60 m.

Let width of the river be DC= DB+BC ⇒ ) m.

In ∆ABC

tan 60° =

20 = √3

-----(1)

Now in ∆ABD

tan 30° =

Therefore width of river = ) m.

⇒

⇒ ⇒ ⇒ m.

Width of river is m.

Let the height of tower is BD = h (m.)

Now in ∆ABC

tan 45° = ⇒

1 =

7+h = ------(1)

Now in ∆DBC

tan 30° =

=

----------(2)

On substituting value of in eqn.(1)

7+h = √3h

√3h- h = 7

h (√3-1) = 7

h = ⇒

= 9.56m.

Therefore height of tower is 9.56m.

Let the height of tower is AB = h (m.)

Now in ∆ABD

tan 45° =

1 =

h = ------(1)

Now in ∆ABC

tan 30° =

=

=

√3h = 2 -------(2)

On substituting value of y from eqn. (1) in eqn. (2)

√3h = 2

√3h = 2

√3h-h = 2

h (√3-1) = 2

h =

on rationalsing above fraction we get,

h =

h =

Therefore height of tower is m.

In fig.BD is the height of the tree. Let the broken part touches the ground at point C

BD = AB+AD

in ∆ABC

tan 30° =

=

√3h = 10

h =

h = m. -------(1)

Again ∆ABC.

COS 30° =

√3

=

= ------(2)

Therefore height of tree = () m.

Adding eqn. (1) and (2) we get,

+

= ⇒ 10√3

10× 1.732 ⇒ 17.32 m.

Therefore height of tree is 17.32 m.

In the fig. Let AB be the height of the balloon.

In ∆ABC

sin 60° =

=

2h = 215√3

h = ⇒ 186.19 m.

Therefore height of balloon from the ground is 186.19 m.

In fig. AB be the height of the cliff.

Let the distance between the two men is DC.

DC = m.

In ∆ABC

tan 60° =

√3 =

√3 = 80

------(1)

In ∆ABD

tan 30° =

--------(2)

Adding eqn. (1) and (2) we get,

=

⇒ ⇒ ⇒ 184.75 m.

Therefore distance between two men is 184.75 m.

Let the position of the sun be at O and AB be the height of the pole.

Now BC is the shadow cast by the pole. From point C, the angle of elevation of the top of the pole (point A) and the sun would be the same.

Since length of shadow is equal to the height of the vertical pole.

Therefore AB = BC

In ∆ABD

tan θ =

tan θ = ⇒ 1

θ = tan 1 (since tan 45°= 1)

θ = 45°

Therefore angle of elevation is 45°

In the fig let DC is the first pole

of height h (m)

In ∆AED

tan 30° =

=

√3h = 15

h = ⇒ 8.66m

Therefore the height of first

pole is 24 – 8.66 = 15.34m

Let the height of building be 'h'.
Let the distance between P and foot of building is 'x' metres.

In ∆PRS

tan 60° =

√3 =

h = √3x ---------(1)

In ∆QRS

tan 45° =

tan 45° =

1 =

h = 20+ ---------(2)

on substituting the value of h from eqn. (2) in eqn. (1)

20+= √3

√3 – = 20

On rationalising above fraction we get,

⇒ ⇒ 10(√3+1)

x = 10(1.732+1)

Therefore Station P has to send the team. And the distance between station P and the building is 27.32 m.

In the fig. AB is the height of the cliff.

AB = AE+EB ⇒ h+10

In ∆AED

tan 45° =

1 =

DE = h -----(1)

In ∆DEB

tan 30° =

DE = 10√3 -----(2)

From eqn. (1) and eqn. (2) we get,

h = 10√3

Height of cliff = AE+EB

= 10√3+10

= 27.32 m.

Therefore distance of the cliff from the ship is 10√3 and Height of cliff is 27.32m.

In the fig. AB is the height of hill AB = AE+EB ⇒ h+8

In ∆AEB

tan 60° =

√3 =

h = √3 DE ------- (1)

In ∆DEB

tan 30° =

DE = 8√3 ------(2)

From eqn. (1) and eqn. (2) we get,

h = √3× 8√3 ⇒ 24 m.

Height of hill = 24+8 ⇒ 32 m.

On substitution value of h in eqn. (1)

24 = √3 DE

DE =

DE = ⇒

DE = 8√3 m.

Therefore distance between ship and hill is 8√3 m and height of the hill is 32 m.

In fig let the height of the other temple is h (m.) and distance between two temple is x (m.)

In ∆ABC

tan 60° =

tan 60° =

√3 =

x = ⇒

x = (1)

In ∆AED

tan 30° =

=

X = h√3 (2)

From eqn. (1) and eqn. (2) we get,

√3h =

h = = 16.67m

On substituting the value of ‘h’ in eqn (2)

X = ⇒ 28.87m

Therefore height of the temple is 50-16.67 = 33.33m

And the distance between the two temples is 28.87m

In the fig let C be the initial position of the aeroplane. After 15 seconds the position of the aeroplane becomes E.

In ∆ABC

tan 45° =

1 =

Y = 3000m ------(2)

In ∆ADE

tan 30° =

=

x + y = 3000

Using equation (1) to replace value of y, we get

x = 3000√3 – 3000

⇒ 3000(√3-1)

⇒ 2196m

Since the distance travelled by aeroplane in 15 seconds is 2196m. Therefore distance travelled by aeroplane in 1 hour =

⇒ 527.04 km/hr

Therefore speed of aeroplane is 527.04 km/hr

Given: An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°.

To find: the speed of the aeroplane in km / hr.

Solution:

Draw the figure according to given information





In the fig


let C be the initial position of the aeroplane. After 10 seconds the position of the aeroplane becomes E.

In ∆ABC

tan 60° =


we know tan 60° = √3

⇒ √3 =


It is given it is flying horizontally 1 km above the ground.

⇒ √3y = 1000 m

⇒ y = -------(1)

In ∆ADE

tan 30° =



we know tan 30° =

⇒ =

⇒ x + y = 1000√3

On substituting value of y from eqn (1)

x + = 1000√3

x = 1000√3 -

⇒ x =

⇒ x =

⇒ x =


⇒ x = 1154.7m

Since the distance travelled by aeroplane in 10 seconds is 1154.7 m.


As 1 hr = 3600 sec and 1 km = 1000 m

Therefore distance travelled by aeroplane in 1 hour = ⇒ 415.69 km/hr

Therefore speed of aeroplane is 415.69 km/hr

In the fig CE is the height of the pole and x be the distance between tower and pole.

In ∆ADE

tan 45° =

1 =

x = 50-h ----- (1)

In ∆ABC

tan 60° =

√3 =

x = -----(2)

On substituting value of x in eqn (1), we get

50 – h =

h = 50 - ⇒

⇒ 21.13m

Therefore the height of the pole is 21.13m

In the fig let CE be the height of the first tree

and AB is the height of the second tree.

In ∆ADE

tan 45° =

1 =

h = 60m

Therefore height of the first tree is 80-60 = 20m

In the fig let RP be the leaning tree, R & S be the two points at distance ‘a’ and ‘b’ from point Q.

In ∆PQR

tan θ° =

tan θ° =

x = ………….(1)

In ∆PQS

tan α = ⇒

tan α = ……………….(2)

In ∆PQT

tan α = ⇒

tan β = ⇒

tan β = …………….(3)

On substituting value of x from eqn (1) in eqn (2) we get,

tan α =

h tan α + a tan θ tan α = h tan θ

h tan α = tan θ(h-a tan α)

tan θ = …………….(4)

Now on substituting value of x in eqn (3)

tan β =

Now on substituting value of tan θ in eqn (4)

tan β =

h²tan β = 0

h(h tan β = 0

h()+tan αtan β(b-a) = 0

h = Proved

In the fig PQ be the height of the tower.

In ∆QRT

tan 45° =

1 =

h = x ……….(1)

In ∆QRT

tan 60° =

√3 =

√3x = h+40 ………….(2)

On substituting the value of x from eqn (1) in eqn (2)

√3h = h+40

h (√3-1) = 40

h = ⇒ 54.64m

Therefore height of the tower is = h+40 = 54.64+40 ⇒ 94.64m

In the fig B is the position of the cloud and C is the point of reflection of the cloud in the lake.

In the fig BD = x

DQ = AP= 2500m

QC = BQ = BD + DQ = (2500 + x)m

DC = DQ +QC = 2500 + 2500 + x = (5000 + x)m

In ∆QRT

tan 45° =

1 =

AD = 5000 + x …………….(1)

Tan15° = tan (45° - 30°) = [Using formula tan(α-β) = ]

Tan15° = =

In ∆ABD

tan 15° =

=

(√3+1)x = (√3+1)(5000+x)

√3x + x = 5000√3 + √3x – 5000 - x

2x = 5000(√3-1)

x = 2500(√3-1)m

Now BQ = BD + DQ = x + 2500

(On substituting value of x)

BQ = 2500(√3-1) + 2500

⇒ 2500√3 – 2500 + 2500

2500√3m

Therefore height of cloud is 2500√3m

In the fig A is the point of observation and C is the position of the cloud.

Let the distance between the cloud and point of observation is x.

In ∆ACD

Sin α =

CD = AC Sin α ⇒ x Sin α

Cos α =

AD = x Cos α …………………….(1)

CE = CD + DE = (h + x Sin α)

EF = CE = (h + x Sin α)

DF = DE + EF = (h + h + x Sin α) = (2h + x Sin α) ……………..(2)

In ∆ADF

tan β =

On substituting value of DF & AD

from above eqns (1) and (2)

tan β =

=

2h Cos β + x Sin α Cos β = x Sin Cos α

x (Cos α Sin - Sin α Cos β ) = 2h Cos β

x =

On dividing numerator and denominator by Cos α Cos β, we get

x =

Therefore the distance between cloud and point of observation is m

In the fig let B & C be two mile stones. And height of the aeroplane is AD

In ∆ABD

tan α =

tan α =

h = BD tan α ……………………(1)

In ∆ACD

tan β =

tan β =

CD = …………………(2)

On adding eqn (1) and (2)

BD + CD =

Now as B and C are milestones, the distance between them = 1
Therefore, BD + CD = 1

1 =

h =

Proved

In the fig PQ is the post off height ‘a’ and AB is the tower of height ‘h’

In ∆BAP

tan α =

tan α =

AP = h tan α …………………………(1)

In ∆BRQ

tan β =

tan β =

AP = ………………………(2)

From eqn (1) and (2) we get,

=

h tan β = (h-a) tan α

h tan β = h tan α – a tan α

h (tan α - tan β) = a tan α

h =

Therefore height of the tower is

Now AP = =

⇒

⇒

Therefore distance is ⇒

In the fig let the length of ladder is h (m)

In ∆AEB

Sin α =

Sin α =

_{AE = h Sin α}

Cos α =

Cos α =

_{BE = h Cos α}

In ∆DEC

Sin β =

Sin β =

_{DE = h Sin β}

Cos β =

Cos β =

_{CE = h Cos β}

_Now

=

=

=

= Proved

In the fig let CD is the height of tower

CD = a + b

In ∆ABD

Cot β =

Cot β =

x = b Cot β ………………(1)

In ∆ADC

tan α =

tan α =

⇒ CD = b Cot β tan α

Therefore height of the tower is b Cot β tan α

In the fig let DC is the observer of the height 1.5m.

In ∆AED

tan θ =

tan θ =

tan θ = 1

θ = tan^-1 1

θ = 45°

Hence the angle of the observation of

the tower from observer’s eye is 45°

In the fig let the height of the stool from the ground is 1.5m i.e. AL = BM = 1.5m

In ∆ACL

tan 60° =

√3 =

CL = ⇒

⇒ = 0.866m

Again in ∆ACL

Sin 60° =

=

AC = = √3m ⇒ 1.732m

Lengths of steps are GH and EF and steps are put at equal distance such that AP =PR = RL

Now consider ∆AGP and ∆ACL

=

=

GP = 0.2887m

Length of step = GH = GP + PQ + QH = GP + AB + GP (Since QH = GP)

GH = 2GP + AB = 2 × 0.2887 + 0.5 = 1.0774m

Similarly we can say that,

ER = ⇒ = 0.577m

Length of the step is EF = ER + RS + SF

⇒ ER + AB + ER = 2ER + AB

EF = 2× 0.5773 + 0.5 ⇒ 1.6546m

In the fig ‘C’ is the position of the kites. Let the length of second kite string is h.

In ∆ABC

Sin 30° =

=

20 + 2x = 100

x = 40m ………….(1)

In ∆CFD

Sin 45° =

=

h = √2 x …………….(2)

On substituting value

of x from eqn (1) in eqn (2)

h = √2 × 40 ⇒ 40√2 m

Therefore length of string of second kite is 40√2 m

In the fig DC is the tower and AB is the hill.

In ∆DCB

tan 30° =

=

BC = 50√3 m

In ∆ABC

tan 60° =

√3 =

AB = 50√3× √3 m ⇒ 150m

Therefore the distance between tower and hill is 50√3m and height of hill is 150m

In the fig AB is the light house of height h (m)

In ∆ABC

tan 30° =

=

x = 100 - √3 h ……………(1)

In ∆ABD

tan 45° =

1=

x = h …………(2)

On substituting value of x from eqn (2) in eqn (1)

h = 100 - √3 h

h + √3 h = 100

h (1 + √3) = 100

h = = ⇒ 50 (√3 - 1)

Therefore height of the light house is 50 (√3 - 1)m

In the fig AB is the building of height 60m and CD is the lamp post of height h (m)

(i) In ∆ABC

tan 60° =

√3 =

BC =

On multiplying and dividing by √3, we get

BC = ⇒ 20√3

Therefore distance between building

and lamp post is 20√3 m

(ii) In ∆AED

tan 30° =

=

= 20√3

AE = = 20

Therefore height of lamp post is CD = AB-AE ⇒ 60-20 = 40 m

Therefore height of lamp post is 40 m

(iii) The difference between the height of the building and the lamp post is 60-40 = 20m

In the fig AB is the light house of height h (m)

In ∆ADC

tan β =

tan β =

h = y tan β or y = ………………..(1)

In ∆ADB

tan α =

tan α =

h = x tan α or x = …………….(2)

The distance between the two ships is BC = x + y

On adding eqn (1) & (2) we get,

x + y = +

⇒

meters PROVED

In the fig AB is the tower on the highway.

In ∆ABC

tan 30° =

=

x + y = 50√3 ………………..(1)

In ∆ABD

tan 60° =

√3 =

y =

On multiplying and dividing by √3, we get

y = ⇒ ………………(2)

Therefore the distance between

the first car and tower is 28.87m

On substituting value of y

from eqn (2) in eqn (1)

x + = 50√3

x = 50√3 -

⇒

⇒ = = 57.73m

The distance between two cars is 57.73m

The distance between second car and tower is (x + y) = 57.73 + 28.87 = 86.60m

In the fig AB is the Rock and CD is the tower.

AB = AE + EB ⇒ h + 100

In ∆ABC

tan 45° =

1 =

x = 100 + h ……………….(1)

In ∆AEB

tan 30° =

=

√3h= 100 + h

h(√3-1) = 100

h =

On multiplying and dividing by √3 + 1, we get

h = ⇒

⇒ 50 (√3 + 1) = 136.6m

Therefore height of the rock = h +100 ⇒ 136.6 + 100 = 236.6m

In ∆ABD

tan 45° =

1 =

x = 150m ………………..(1)

In ∆ABC

tan 30° =

=

x + y = 150√3 ………………..(2)

On substituting value of x from eqn (1) in eqn (2), we get

150 + y = 150√3

y = 150 (√3 - 1) ⇒ 109.8m

Therefore the distance between two ships is 109.8m

In the fig let AD is the Flag-Staff of height h (m)

In ∆ABC

tan 45° =

1 =

x = 5m

In ∆DBC

tan 60° =

√3 =

h + 5 = 5√3

h = 5√3-5

h = 5(√3-1)

h = 3.66 m

Therefore height of flag-staff is 3.66m

In the fig let AB is the Tower of height h (m).
Since the tower is vertical to the ground.

∠ABC = 90°
In a right-angled triangle, we know,

In ∆ABD

tan α =

tan α =

h = 4 tan α …………(1)

In ∆ABC

tan (90°-α)=

⇒ cot α =

⇒ h = 9 cot β ……….(2)

On multiplying eqn (1) and eqn (2), we get

h × h = 4 tan α × 9 cot α

h² = 36
h = ± 36

⇒ h = 6m

Therefore the height of the tower is 6m.

In the fig let AB is the light house of height h (m)

In ∆ABC

tan 30° =

=

√3h = 200 + x ………………….(1)

In ∆ABD

tan 45° =

1 =

h = x ………………(2)

From eqn (1) and (2) we get

√3h = 200 + h

h(√3 + 1) = 200

h =

On multiplying and dividing by √3-1, we get

h =

h = ⇒ 273.2m

Therefore height of the light house is 273.2m

In the fig AB is the light house of height of 200m.

In ∆AED

tan 45° =

1 =

x = 200m ………………….(1)

In ∆ABC

tan 60° =

√3 =

√3y = 200

y =

On multiplying and dividing by √3, we get

y = ⇒ m

Therefore the distance between two ships is:

DC = x + y

x + y = 200 +

⇒

⇒ = 315.47m

Therefore the distance between two ships is 315.47m

In the fig AB is the Chimney and CD is the tower of 40m

In ∆AED

tan 60° =

√3 =

h = √3DE ……………..(1)

In ∆DEB

tan 30° =

=

DE = 40√3 ………………..(2)

On substituting value of DE from

eqn (2) in eqn (2), we get

h = √3 × 40√3 = 120

Therefore height of Chimney is 40 + 120 = 160m

Yes, the height of the chimney meets the pollution norms.

Chimneys are made tall so that smoke should go high in the atmosphere in order to minimize air pollution.

In the fig AD is the position of the aeroplane. Let the width of the river is DC = DB + BC

In ∆ABD

tan 45° =

1 =

x = 210m …………………(1)

In ∆ABC

tan 60° =

√3 =

√3y = 210

y =

On multiplying and dividing by √3, we get

y = ⇒ = 70√3 ……………………(2)

Therefore width of the river is = 210 +70√3 = 331.24m

Let BC be the height of tower which is 10m

Sun`s altitude is 45°

Let AB would be the shadow of the tower x meters

∴ tan 45° = BC /AB (tan θ = perdendicular/base)

⇒ 1= 10/x

x= 10m

The ratio of the length of rod and its shadow = 1: √3

Let the angle of elevation of sun be θ

tan θ = P / B (P = perpendicular, B = base)

Here tan θ = 1: √3 =

tan θ = (by rationalizing the denominator)

θ = 30° (∵ tan 30° = )

Here, angle of elevation = 60°

Distance between the foot of tower and the shadow = 100m

Let height of the tower be h meters

Angle of elevation = tan θ =

tan 60° =

√3 × 100 = h

Height of tower = 100 √3 m

The ratio of height of a tower and the length of its shadow= √3 :1

Angel of elevation = θ

tan θ = √3 :1

tan θ = √3 (∵ tan 60° = √3)

∴ θ = 60°

Here length of the shadow of a vertical pole is equal to its height

Let them both be x m

Angle of elevation = tan θ = P/B (p = perpendicular, b= base)

Here p = b = x

∴ tan θ = x/x

tan θ = 1

∴ θ = 45° (∵ tan 45° = 1)

Altitude of the sun = 60°

Length of the vertical tower = 30 m

Height of tower be h meters

tan θ = H/ B

tan 60° =

h = 30 √3 m

Given BC = a and BD = b
Let the height AB = h
In right ΔABD,

tan θ =

tan 30° =

……………1

In Δ ABC

tan 60° =

√3 a = h………………2

Multiplying 1 and 2

h² = b × √ 3 a

h = √ab

Let the height of the tower be AC = x m

Distance from the foot of the vertical tower = 20m

tan 60° = AC/ BC

√3 = x / 20 ( tan 60° = √3)

x = 20√3

Since the angles of elevation are complementary then if one angle is θ other would be 90° - θ

Here CD is the height of tower which forms two complementary angles θ and 90° - θ from its top to the distance a meters and b meters respectively.

In Δ CAD

1

In Δ CBD

Putting value of tan θ From 1

(CD)² = ab

CD = √ab

Let the height of the tower be h meters

Given, the angles of elevation of the top of a tower from two points are complimentary

∴ ∠ACB = θ and ∠ADB = 90° - θ

In Δ ABC

tan θ = 4 / h

h = 4tan θ…………1

In ΔABD

tan (90° - θ) =

h = 9 (cot θ) ………………..( tan (90° - θ) = cot θ ) 2

cot θ = h/9

cot θ =

1/tan θ =

9 = 4 tan²θ

tan θ = 3/2

Height of tower (h)= 4 × 3/2………..putting value of tan θ in 1

= 6m

Here AB is the light house of height h meters

The angles of depression from the light house are 30° and 45°

The distance between the two ships is CD

In Δ ABC

tan 45° =

⇒ BC = AB (∵ tan 45° = 1)

⇒ BC = h

In Δ ABD

tan 30° =

⇒ (h + CD) = √3 h (AB = BC = h)

⇒ CD = (√3 -1) h

Let x be the angle of depression of object A from the point O₂

And y be the angle of depression of object A from the point O₁

In Δ AO₁C

∠O₁AC + ∠AC O₁ + ∠ACO₁ = 180° (angle sum property)

∠O₁AC + 90° + 60° = 180°

∠O₁AC = 30°

Through O₁ , draw O₁M ∥ AC

And through O₂ draw O₂N∥ AC

Now O₁M ∥ AC and AO₁ is transversal

∴ ∠O₁AC = y = 30° = ∠MO₁A (vertically opposite ∠s)

Similarly, O₂N∥ AC and AO₂ is transversal

∠NO₂ A = x = 45° = ∠O₂ AB (vertically opposite angles)

∴ Angles of depression are 30° and 45°

Given: The angle of elevation of the top of a tower standing on a horizontal plane from a point C is α. After walking a distance d towards the foot of the tower the angle of elevation is found to be β.

To find: The height of the tower

Solution:

Let h be the height of the tower on horizontal plane.

Let α be the angle of elevation from point C and β be the angle of elevation from point B

Given CB = d

In Δ PCB

tan (α) =

x =

In Δ CDB

tan(β) =

⇒ tanβ ( h - d tanα) = h tanα

⇒ h tanβ - d tan a tanβ = h tan a

⇒ h (tanβ – tan a) = d tan a tanβ





Use the formula:

Hence (b) is the answer.

In ΔABE

tan 30° = x/a

⇒ x = a tan 30°

Now, in Δ ECD

tan 60° = y/a

⇒ y = a tan 60°

⇒ x :y = 1:3

Let h be the height of the tower

∴ tan 30°= h/ AB (tan θ = perpendicular / base)

………..1

When the height is tripled h becomes 3h

tan θ = 3h / AB

tan θ = (from 1)

tan θ =

tan θ = √3 ( by rationalizing the denominator)

θ = 60° (tan 60° = √3)

Here ED = 14m and AC = 20 m are two poles

⇒ AB = 20 – 14 = 6 m

Wire AE connects them making angle with horizontal of 30°

We have to find AE which is length of wire

Sin 30° =

=

AE = 12 m

Here AB = 25 m is cliff and CE = (25 + x) m is tower

In Δ ADE

tan θ = …………….1

In Δ ABC

tan θ = …………….. 2

From 1 and 2

x =25 m

Total height of tower is 25 + x = 25 +25 = 50 m

Here AD = h meter is the tower

The ships B and C are 100 m apart so BC = 100 m

In Δ ACD

tan 45° =

CD = h (∵ tan 45° = 1)

In Δ ABD

tan 30° =

=

h =

h= 100 + h

h = m

On rationalizing the denominator

h =

h = 50 (m

A is the position of the cloud, B is the point 200 m above the lake and F is the reflection in the lake.

Here AE = EF

EF = (m + 200) m

In ΔABC

BC = m cot 30° ……………….1

In Δ BCF

CF = 200 + m +200

= (400 + m) meters

BC = (400 + m) cot 60° ………………….2

From 1 and 2

m cot 30° = (400 + m) cot 60°

m =

3 m = (400 + m)

m = 200 m

The height above the cloud above the lake is (200 + 200)m = 400 m.

Let AB be the tower of height 100 m

BC is the total distance of shadow formed at two different angles namely ACB and ADB 30° and 45° respectively

In Δ ACB

tan 30° =

= 100

a = 100

In Δ ADB

tan 45° =

100√3-x = 100

x = 100(√3-1) m

Let AB be the line joining the distance between two persons. Given C is the midpoint of AB so AC = a/2 and CB = a/2

Height of taller person is double the height of shorter person which is 2h and h respectively forming complementary angles

tan x = ……………..1

And tan 90- x =

cot x= ………………2

Multiplying 1 and 2

tan x cot x = ×

a^{2 =} 8 h²

a = 2√2 h

Let A and B be the position of the cloud and its reflection in the lake.

Let the height of the cloud be H m.

Given EF = h m, ∠AED = θ and ∠DEB = 45°

As EF || CD

CD = h m

By law of reflection,

AC = BC = H

AD = AC – DC

= H – h

And,

BD = BC + DC

= H + h
In ΔDEB,


⇒ BD = DE
⇒ DE = H + h ..... (1)
In ΔAED,


...... (2)
From (1) and (2),

⇒ H - h = (H + h) tan θ
⇒ H - h = H tan θ + h tan θ
⇒ H - H tan θ = h + h tan θ
⇒ H (1 - tan θ) = h (1 + tan θ)


As we know,

⇒ H = h (tan 45° + θ)

Let the AB be the height be the tower

Let D be the point where the tower subtends angle of 30°

Let C be the point where such that CD = h meters. From C the angle of depression subtended at the foot of the tower is 60°

In Δ CDB

tan 60° =

BD = h cot 60°

BD = …………….1

In Δ ADB

tan 30° = AB/ BD

= AB / BD

AB = ×

AB = m

Here AB is the chimney of height h

By walking x meters toward chimney the angle of elevation changes from 30° to 60°

In Δ ABH

tan 60° =

h = √3 y

= y …………….1

In Δ ABG

tan 30° =

=

√3 h = x+y

√3 h - = x (from 1)

= x

h =

In Δ DBC

tan 45° = h /y

h = y (∵ tan 45° = 1)……….1

In ΔACD

tan 30° =

=

√3 h = 2x + h (from 1)

2x= (√3- 1) h
h =

h = (√3 +1)x meter

Let AB and CD be the two poles of height h meters and 2h meters respectively such that BD be a km i.e.; the distance between the two poles and P be the midpoint of BD .

Given ∠APB = θ and ∠CPD = 90 – θ

In Δ ABP

tan θ =

tan θ = …………………1

In Δ CDP

cot (90° – θ) = PD / CD =

cot (90° – θ )=

tan θ = ………….2

Equating 1 and 2

=

8 h ² = a²

h = m

Let AE be the wire to connect the two poles ED and AC of height 10m and 16m forming angle of 30° with horizontal

In Δ AEB

Sin 30° =

=

AE = 12 m

Let AB be the tower of h meters and CD be the girl of 1.6m height casting shadow = DE of 4.8 m, standing at the distance of 3.2m from the tower.

In Δ CDE

tan ∠CED = CD/ DE = ………………….1

tan ∠CED = 1/3

In Δ ABE

tan ∠CED = = ………………….2

From 1 and 2

=

h = m

h = 2.5m

AB is the height of the tower and BC is the length of the shadow forming angle of elevation

tan θ =

Given AB= h and BC = √3 h

tan θ =

tan θ =

θ = 30°

Here AB is the tower of height 75m forming angle of depression = 30°

∴ Angle of elevation is 90° - 30° = 60°

tan 60° =

√3 =

x = ×

x =25 √3

Given AB be the ladder of 15 m length and AC be the wall of h meters making angle with wall as 60°.

Cos 60° = =

=

AC = m

Here AB is the tower of 150m which forms the angel of depression = 30° from the top

Hence the angle of elevation from the car is 90° - 30° = 60°

In Δ ABC

tan 60° =

BC =

BC = 50 √3 m

Consider the height of tower be √3 h
∴ height of shadow =h .
In a triangle ABC,

tan ∠ACB =

tan ∠ACB = √3

∠ACB = 60°.
Therefore, angle of elevation is 60° (∵ tan 60° = √3)

Let AB be the tower of h meters height forming angle of elevation= 45° from 50m distance from the tower

In Δ ABC

tan 45° =

h = 50 m

Here AC is the ladder of x meters placed against the wall of h meters at the distance of 2m from the wall.

It forms the angle of elevation to the top of wall as 60°

Cos 60° =

=

Hypotenuse = 4 (length of the ladder)