Question

plz give me the answer step by step​

Answer 1

Answer:

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Step-by-step explanation:

Given that AB is the tower

P, Q are the point at distance 4m and  9m respectively

Also, PB = 4m , QB = 9m

& Angle of elevation from P is a

Angle of elevation from Q is ß.

Given a and B are supplementary.

a + B = 90°

We need to prove AB = 6m

In △ ABP

tan a = $\frac{AB}{BP}$ ------ (1)

In △ABQ

tan B = $\frac{AB}{BQ}$

tan (90 - a) = $\frac{AB}{BQ\\}$ ------ (given a + B = 90° )

cot a  = $\frac{AB}{BQ}$ --------(tan (90-theta)=cot theta)

$\frac{1}{tan a}$ = $\frac{AB}{BQ}$-----------(cot theta = 1/tan theta)

tan a = $\frac{BQ}{AB}$-------(2)

From (1) and (2)

AB/BP  = BQ/AB

AB X AB = BQ X BP

(AB)^2 = 4 * 9

(AB)^2 = 36

AB = (36)

AB = 6m

Hence, height of the tower is 6m.

Hence proved.