This page 10th samacheer kalvi math solution for exercise 7.2 part 7 is going to provide you solution for every problems that you find in the exercise no 7.2

**Question 17 :**

From the top and foot of a 40 m high tower, the angles of elevation of the top of a lighthouse are found to be 30° and 60° respectively. Find the height of the lighthouse. Also find the distance of the top of the lighthouse from the foot of the tower.

**Solution :**

AB = x m and BD = 40 m

In triangle ABC :

∠ACB = 30°

tan θ = opposite side/Adjacent side

tan 30° = AB/BC

1/√3 = x/BC

BC = x√3 ----(1)

In triangle ADE :

∠AED = 60°

tan θ = opposite side/Adjacent side

tan 60° = AD/DE

√3 = (x + 40)/DE

DE = (x + 40)/√3 ---->(2)

(1) = (2)

x√3 = (x + 40)/√3

3 x = x + 40

3x - x = 40

2x = 40

x = 20

Height of the tower = 40 + 20 = 60 m

**Question 18 :**

The angle of elevation of a hovering helicopter as seen from a point 45 m above a lake is 30° and the angle of depression of its reflection in the lake, as seen from the same point and at the same time, is 60°. Find the distance of the helicopter from the surface of the lake.

**Solution :**

From the given information, we can draw a rough diagram

In a right triangle FAE :

tan θ = opposite side/Adjacent side

tan 30° = FE/AE

1/√3 = (h-45)/AE

AE = (h - 45) √3 ----(1)

In right triangle ACE :

tan θ = opposite side/Adjacent side

tan 60° = EC/AE

√3 = (ED+DC)/AE

√3 = (45+h)/AE

AE = (45 + h)/√3----(2)

(1) = (2) (BC = AD)

(h - 45) √3 = (45 + h)/√3

3(h - 45) = 45 + h

3h - 135 = 45 + h

2h = 45 + 135

2h = 180

h = 90

Hence the distance of the helicopter from the surface of the lake is 90 m.

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