Question

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Answer 1

Answer:

Hi your answer is as follows

Step-by-step explanation:

Given AB is the tower.

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.

Given that α and β are supplementary. Thus, α + β = 90

In triangle ABP,

tan α = AB/BP – (i)

In triangle ABQ,

tan β = AB/BQ

tan (90 – α) = AB/BQ (Since, α + β = 90)

cot α = AB/BQ

1/tan α = AB/BQ

So, tan α = BQ/AB – (ii)

From (i) and (ii)

AB/BP = BQ/AB

AB^2 = BQ x BP

AB^2 = 4 x 9

AB^2 = 36

Therefore, AB = 6.

Hence, height of tower is 6m.

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Answer 2

$let \: \angle APB = \theta \\ then \angle \: AQB= ( 90 \degree - \theta) \\ in \: right \ triangle \: APB \\ \tan \theta = \frac{AB}{PB} \\ tan \theta = \frac{AB}{9}...................(1) \\ then \: in \: right \: triangle \: \: ABQ\\ tan(90 \degree - \theta) = \frac{AB}{QB} \\ cot \theta = \frac{AB}{4} ............(2) \\ multiplying \: (1)and(2) \\ \frac{AB}{9} \times \frac{AB}{4 } = tan \theta \times cot \theta \\ \frac{AB} {36}^{2} {} = 1 \\ {AB}^{2} = 36 \\ AB = \sqrt{36} \\ AB= 6m$

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