Question

Solve this question
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Answer 1

here is your answer friend

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

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Answer 2

above answer is totally correct