This page 10th samacheer kalvi math solution for exercise 7.2 part 3 is going to provide you solution for every problems that you find in the exercise no 7.2

**Question 7 :**

A lamp-post stands at the centre of a circular park. Let P and Q be two points on the boundary such that PQ subtends an angle 90° at the foot of the lamp-post and the angle of elevation of the top of the lamp post from P is 30° . If PQ = 30 m, then find the height of the lamp post.

**Solution :**

Let O be the centre of the park and OR be the lamp post. P and Q are two points on the boundary of the circular park. Given that PQ = 30m,

∠POQ = 90°

In a right triangle OPQ, ∠POQ = 90°

OP = OQ = radius. So ∠OPQ = ∠OPQ = 45°

OP = PQ x cos 45°

OP = 30/√2 ==> (30/√2) x (√2/√2) ==> 15 √2

In a triangle RPO

tan 30° = OR/OP

1/√3 = OR/15√2

OR = (15√2/√3) ==> (15√2/√3) x (√3/√3)

= 15√6/3 ==> 5√6

Hence the height of the lamp post is 5√6 m

**Question 8 :**

A person in an helicopter flying at a height of 700 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30° and 45°. Find the width of the river. (√3 = 1.732 )

**Solution :**

From the given information, we can draw a rough diagram

AD = 700 m

In triangle ACD, we need to find the length of DC.

∠ACD = 45°

tan θ = opposite side/Adjacent side

tan 45° = AD/DC

1 = 700/DC ==> DC = 700 --(1)

In triangle ABD, we need to find the length of BD.

∠ABD = 30°

tan θ = opposite side/Adjacent side

tan 30° = AD/BD

1/√3 = 700/BD ==> BD = 700√3 --(2)

(1) + (2)

Width of the river = BD + DC ==> 700 + 700√3

= 700(1 + √3) ==> 700 (1 + 1.732) = 700 (2.732)

= 1912.4 m

Width of the river is 1912.4 m

**Question 9 :**

A person X standing on a horizontal plane, observes a bird flying at a distance of 100 m from him at an angle of elevation of 30°. Another person Y standing on the roof of a 20 m high building, observes the bird at the same time at an angle of elevation of 45°. If X and Y are on the opposite sides of the bird, then find the distance of the bird from Y.

**Solution : **

In triangle ABC

∠ABC = 30°

sin θ = opposite side/Hypotenuse side

sin 30° = (AE + EC)/BC

1/2 = (x + 20)/100

100 = 2 (x + 20)

100 = 2x + 40

100 - 40 = 2x ==> 2x = 60 ==> x = 30

In triangle AYE

∠AYE = 45°

sin 45° = AE /EY

1/√2 = x / EY

1/√2 = 30 / EY

EY = 30√2

Hence the required distance of the bird from Y is 30√2 m.

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