PLeAse AnSwEr ThiS QuEsTiON ❣️❣️

1.	PLeAse AnSwEr ThiS QuEsTiON ❣️❣️
Answer» Answer: [Figure in the attachment] Given – A point above the lake is Q. The DISTANCE between the point P and Q is 200 m. ANGLE of elevation from the point Q to the cloud is 30°. The angle of depression from the point Q to the reflection of the cloud in the lake is 60°. To Find – Height of the cloud. Solution – We know that the length PQ is 200m. Hence, AB = 200 m. Now, height of cloud = AB + BC. We know the length of AB, We need to find BC. In ∆ QBC, tan ∅ = Perpendicular/Base Hence, tan 30° = BC/QB. ➸ tan 30° = X/QB [Assumed BC as x] ➸ 1/√3 = x/QB ➸ QB = √3 x ....k1) In ∆ QBD, tan ∅ = Perpendicular/Base Hence, tan 60° = BD/QB ➸ √3 = (200 + AD)/QB ➸QB√3 = 200 + AD ➸ QB = (200 + AD)/√3 ....(2) Equating equation 1 and 2, ➸ QB = √3 x = (200 + AD)/√3 ➸ 3 x = 200 + AD [AD = 200 + x] ➸ 3 x = 200 + 200 + x ➸ 3 x = 400 + x ➸ 3 x - x = 400 ➸ 2 x = 400 ➸ x = 400/2 ➸ x = 200 Hence, Height of cloud = 200 + x = 200 + 200 = 400 m.

Answer»

Answer:

[Figure in the attachment]

Given –

A point above the lake is Q.
The DISTANCE between the point P and Q is 200 m.
ANGLE of elevation from the point Q to the cloud is 30°.
The angle of depression from the point Q to the reflection of the cloud in the lake is 60°.

To Find –

Solution –

We know that the length PQ is 200m. Hence, AB = 200 m.

Now, height of cloud = AB + BC.

We know the length of AB, We need to find BC.

In ∆ QBC,

tan ∅ = Perpendicular/Base

Hence, tan 30° = BC/QB.

➸ tan 30° = X/QB

[Assumed BC as x]

➸ 1/√3 = x/QB

➸ QB = √3 x ....k1)

In ∆ QBD,

tan ∅ = Perpendicular/Base

Hence, tan 60° = BD/QB

➸ √3 = (200 + AD)/QB

➸QB√3 = 200 + AD

➸ QB = (200 + AD)/√3 ....(2)

Equating equation 1 and 2,

➸ QB = √3 x = (200 + AD)/√3

➸ 3 x = 200 + AD

[AD = 200 + x]

➸ 3 x = 200 + 200 + x

➸ 3 x = 400 + x

➸ 3 x - x = 400

➸ 2 x = 400

➸ x = 400/2

➸ x = 200

Hence, Height of cloud = 200 + x = 200 + 200 = 400 m.

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