# Examples II

On the same side of a tower two objects are observed from the top of the tower. Their angles of depression are 45Â° and 60Â°. Find the distance between the objects, if the height of the tower is 300 m.

Let DC be the height of the tower and A, B be the positions of the two objects.Â Â

Let the distance between the objects be x m.

In theÂ Î”Â BCD, tan 60Â°=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 =Â Â mÂ (1)

In theÂ Î”Â ACD,

Â Â Â Â Â Â tan 45Â° =Â

â‡’Â Â Â Â Â Â Â Â Â Â Â 1 =Â

â‡’Â Â x + BC = 300

â‡’Â x +Â Â = 300

â‡’Â Â Â Â x = 300 -Â Â = 126.79 m

Hence, the distance between the two objects isÂ 126.79 m.

Â

Two vertical poles are fixed 60 m apart. The angle of elevation of the top of the first as seen from the top of the second, which is 160 m high, is 30Â°. Find the height of the first pole.

Let AP and CQ be the two vertical poles. Let the height of the first pole AP be h m.

Given: CQ = 160 m , PQ = 60 m. Â

In theÂ Î”ABC, tan 30Â°=Â

â‡’Â Â Â Â Â Â Â Â Â Â Â Â Â Â

â‡’Â Â Â Â (160-h) = 60

â‡’Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â h = 160 -Â

âˆ´Â Height of the first pole = 125.36 m.

A vertical stick 10 cm long, casts a shadow 6 cm long on the ground. Under similar conditions, a tower casts a shadow 10 m long. Find the height of the tower.

Let CB = 10 cm be the height of the stick and EF = h m be the height of the tower.

Given:

DE = 10 m, AB = 6 cm

In

Î”Â ABC,tanÂ Î¸Â =Â

In theÂ Î”Â DEF,

tanÂ Î¸Â =Â

â‡’Â Â Â Â Â Â Â Â â‡’Â Â h =Â Â = 16.67 m

The required height of the tower is 16.67 m.

The pilot of a helicopter, at an altitude of 1200 m finds that two ships are sailing toward it in the same direction. The angles of depression of the ships as observed from the helicopter are 60Â° and 45Â° respectively. Find the distance between the ships.

Let CD be the height at which the helicopter is flying CD = 1200 m.

Let A and B the position of the two ships.Â

In theÂ Î”BCD, tan 60Â° =Ââ‡’Â

â‡’Â BC =Â Â mÂ Â Â Â ...Â Â Â (1)

In theÂ Î”ACD,

Â Â Â Â Â Â Â Â Â tan 45Â° =Â

â‡’Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1 =Â

â‡’Â Â Â Â Â Â Â Â x + BC = 1200

Substituting for BC from (1)

â‡’Â Â Â x +Â Â = 1200

â‡’Â Â Â Â Â Â Â Â Â Â Â Â Â Â x = 1200 -Â Â = 507.16 m

Hence, the distance between the ships is 507.16 m.

A tree 12 m high is broken by the wind in such a way that its top touches the ground and makes an angle of 60Â° with the ground. At what height from the bottom, the tree is broken by the wind?

Let ACB be the total length of the tree broken AC + CB = 12 m.Â

Let BC (h) be the height from the bottom of the tree, broken by the wind.

CD = CA = 12 - h

In the triangle ABC,

sin 60Â° =Â

â‡’Â

â‡’Â 2h = 12-h

â‡’Â h =Â Â = 5.57 m

Hence, the tree is broken at a height of 5.57 from the bottom.Â

Â

Â

A surveyor wants to find out the height of a hill. He observes that the angle of elevation of the top of the hill at points A and B, 300 m apart, lying on a straight line, passing through the base of the top on the same side of the hill, are 30Â° and 45Â° respectively. What is the height of the hill?

Let the height of the hill be h m.

**Given:**

AB = 300 mÂ

In theÂ Î”BCD,

Â Â Â Â Â Â tan 45Â° =Â

â‡’Â Â Â Â Â Â Â Â Â Â Â 1 =Â

â‡’Â Â Â Â Â Â Â Â Â BC = h

In theÂ Î”ACD,

tan 30Â°=Â

â‡’Â Â Â Â Â Â Â Â

â‡’Â Â Â Â Â 300 + BC =Â Â h

â‡’Â Â Â 300+h=Â Â h

h =Â Â = 409.84 m

Hence, the height of the hill is 409.84 m.

Â

Â

The string of a kite is 250 m long and it makes an angle of 60Â° with the horizontal. Find the height of the kiteÂ above the ground (Assuming the string to be taut).

Let BC be the height of the kite in h m.

AC = 250 m is the length of the taut string.Â

In theÂ Î”ACB,

sin 60Â° =

â‡’Â Â

â‡’Â h =Â Â x 250 = 216.5 m

The required height of the kite from the ground isÂ 216.5 m.

Â