# Examples III

The angles of depression of two ships from the top of a light house are 45Â° and 30Â° towards east. If the ships are 200 m apart, find the height of the light house.

Let AB be the height of the light house be h metres and let C and D be the positions of the two ships.

**Given:**Â CD = 200 m.

In theÂ Î”ABC, tan 45Â° =Â

Â

â‡’Â 1 =Â

â‡’Â BC = h

In theÂ Î”ABD,

Â tan 30Â° =Â

â‡’Â Â Â

â‡’Â Â Â

â‡’Â Â Â h = h + 200

â‡’Â Â Â (-1)h = 200

â‡’Â Â Â h =Â Â = 273.22 m

Hence, the required height of the light house is 273.22 m.

Â

From the top of a building 60 m high, the angles of depression of the top and the bottom of a vertical lamp post are observed to be 30Â° and 60Â° respectively. Find,Â

(a) The horizontal distance between the lamp postÂ and the building.

(b) The difference between the heights of the buildingÂ and the lamp post.

(a) Let BD be the height of the building = 60 m. Let AB be (x m) the distance between the lamp post and theÂ building.

In theÂ Î”ABD,

tan 60Â° =Â

â‡’Â Â Â =Â

â‡’Â x =Â Â = 34.64 mÂ .......(1)

(b) Let the height of the lamp post be h m.

In theÂ Î”ECD,

tan 30Â° =Â

â‡’Â Â Â Â Â Â Â Â Â Â

â‡’Â Â Â Â Â Â Â Â Â Â x =Â Â (60-h)

â‡’Â Â Â Â Â Â Â Â Â Â Â Â =Â (60-h)Â Â Â Â Â Â [QÂ x =Â Â from (1)]

â‡’Â Â Â Â Â Â Â Â Â Â 20 = 60 - h

â‡’Â Â Â Â Â Â Â Â Â Â h = 40.

The difference between the heightsÂ = 60 - 40Â = 20 m.

Â

An aeroplane at an altitude of 200 m, observes the angles of depression of the opposite points on the two banks of a river to be 45Â° and 60Â°. Find the width of the river.

LetÂ BD = 200 m be the height of the aeroplane and AC width of the river.

In theÂ Î”ABD, tan 60Â° =Â

â‡’Â

â‡’Â AB =Â Â m

InÂ Î”DBC,

tan 45Â° =Â

â‡’Â 1 =Â

â‡’Â BC = 200 m

The width of the river

= AC

= AB + BC

=Â Â + 200 = 315.47 m.

Â

From the top of a hill, the angles of a depression of the two consecutive kilometre stones due last, are found to be 30Â° and 45Â°. Find the height of the hill.Â Â Â

Let AB = h km be the height of the hill.Â

In the

tan 45Â° =Â

â‡’Â Â Â Â Â Â Â Â 1 =Â Â

â‡’Â Â Â Â Â Â Â Â h = BC

In theÂ Î”ABD,

tan 30Â° =Â

â‡’Â Â Â Â Â Â

â‡’Â Â Â Â h = h + 1

â‡’Â Â Â Â (Â - 1)h = 1

â‡’Â h =Â Â = 1.37 km

Â

The angle of elevation of a tower at a point is 45Â°. After going 40 m towards the foot of the tower the angle of elevation becomes 60Â°. Find the height of the tower.Â

Let AB (h) be the height of the tower, D and C be the positions of the person.

Angles of elevation areÂ 45Â° and 60Â°.

In theÂ Î”ABC,

tan 60Â° =Â

â‡’Â Â

â‡’Â BC =Â

In theÂ Î”ABD,

tan 45Â° =Â

â‡’Â 1 =Â

â‡’Â Â + 40 = h

â‡’Â hÂ = 40

â‡’Â Â h =Â

Â Â Â Â Â = 94.65 m

âˆ´The height of the tower is 94.65 m.

Â

Â

The horizontal distance between two trees of different heights is 60 m. The angle of depression of the first tree when seen from the top of second tree is 45Â°. If the height of the second tree is 80 m find the height of the first tree.

Let AB (h) be the height of the first tree in metres and CE = 80 m be the heightÂ of the second tree such that the distance between the trees = BC = 60 m.

Â In theÂ Î”ADE,

tan 45Â°=Â

â‡’Â 1 =Â

â‡’Â 60 = 80 - h

â‡’Â h = 20 m

Hence, the height of the first tree is 20 m.

Â

The angle of elevation of the top of the tower, from two points P and Q at distances a and b respectively from the base and in the same straight line with it, are complementary. Prove that the height of the tower isÂ Â .

Let AB (h) be the height of the tower in metres.

Let the angle of elevation from P and Q beÂ aÂ Â and 90Â°-Î±Â respectively.

BP = a,Â Â Â BQ = b

In theÂ Î”ABP,

Â tanÂ Î±Â =Â Â

â‡’Â tanÂ Î±Â =Â

â‡’Â h = a tanÂ Î±

In theÂ Î”ABQ

Â tan (90Â°-Î±) =Â

â‡’Â Â cotÂ Î±Â =Â

â‡’Â Â h = b cotÂ Î±

Multiplying (i) and (ii),

h^{2Â }= (a tanÂ Î±) (b cotÂ Î±) = ab

â‡’Â h =Â

Â

The angles of elevation of a cliff from a fixed point A isÂ Î¸. After going up a distance of k metres towards the top of the cliff at an angle ofÂ Ï†Â it is found that the angle of elevation isÂ Î±. Show that the height of the cliff, in metres isÂ kÂ .

Let PQ (h) be the height of the cliff in h metres.

AB = k

In theÂ Î”ACB,Â Â = cosÂ Ï†

â‡’Â AC = AB cosÂ Ï†Â Â

Â Â Â Â Â Â = k cosÂ Ï†*Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â *(i)

â‡’Â Â Â = sinÂ Ï†*Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â *

â‡’Â Â BC = AB sinÂ Ï†Â = k sinÂ Ï†Â Â Â Â Â Â Â Â Â Â Â Â (ii)

â‡’Â Â PR = k sinÂ Ï†

In theÂ Î”APQ,Â Â = tanÂ Î¸

â‡’Â Â Â = tanÂ Î¸

â‡’Â Â AP = h cotÂ Î¸Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (iii)

In theÂ Î”BRQ,Â Â Â = tanÂ Î±Â Â Â Â Â Â Â Â Â

â‡’Â Â = tanÂ Î±Â Â Â Â Â (since PR = BC)

â‡’Â Â Â Â Â Â (h - PR) cotÂ Î±Â = AP - AC

â‡’Â Â (h- k sinÂ Ï†Â )cotÂ Î±Â = h cotÂ Î¸Â - k cosÂ Ï†Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (i), (ii) and (iii)]Â

â‡’Â Â Â Â h[cotÂ Î±Â - cotÂ Î¸ ]Â =Â k sinÂ Ï†Â cotÂ Î±Â - k cosÂ Î±

â‡’Â Â Â Â h[cotÂ Î¸-cotÂ Î±] = k [cotÂ Ï†*Â *- sinÂ Î¸Â cosÂ Î±]

h = kÂ .

Â

A round balloon of radius 'a' subtends an angleÂ Î¸Â at the eye of the observer while the angle of elevation of its centre isÂ Ï†Â . Prove that the height of the centre of the balloon, is a sinÂ Ï†Â cosec()

âˆ AOB =Â Î¸Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

â‡’Â âˆ BOC =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Let the height of the centre of the balloon be

CD = h

âˆ COD =Â Ï†Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

BC = AC = a

In theÂ Î”OBC,

â‡’Â Â OC = a cosecÂ .......(1)

InÂ Â Î”ODC,

sinÂ Ï†Â =Â

â‡’Â CDÂ = OC sinÂ Ï†

Substitute for OC from (i)

â‡’Â h = a cosecÂ sinÂ Ï†

Â