Question

The angles of elevation of the top of a tower from two points at a distance of 4 m and
16.
9 m from the base of the tower and in the same straight line with it are complementary.
Prove that the height of the tower is 6 m.​

Answer 1

Answer:

P and Q are the points at distance of 4m and 9m respectively. From fig, PB = 4m, QB = 9m. Let angle of elevation from P be α and angle of elevation from Q be β. Therefore, AB = 6.

Step-by-step explanation:

go through the attached 2 sheets

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Answer 2

Step-by-step explanation:

The given situation can be represented as,

Let height of the tower be h m.

Given, the angles of elevation of the top of tower from the two points are complementary.

∴ ∠ACB = θ and ∠ADB = 90 – θ

In ∆ABC,

In ∆ABD,

∴ Height of the tower = h = 4 tan θ = 4 × = 6 m (Using (1))

Thus, the height of the tower is 6 m.