Math, asked by vansh1612, 1 year ago

P is the top and Q the foot of a tower standing on a horizontal plane. A and B are two points on this plane such that AB is 32m and angle QAB is a right angle. it is found that cot(anglePAQ)=2/5 and cot(anglePBQ)=3/5 then find the height of tower (use √5=2.235)

Answers

Answered by 16Devesh16
5
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Answered by aquialaska
4

Answer:

Height of tower is 160 m.

Step-by-step explanation:

Given: PQ is tower

           AB = 30 m

To find: Height of tower

In ΔPAQ,

using trigonometric ratio we get

cot\,PAQ=\frac{AQ}{PQ}

\frac{2}{5}=\frac{AQ}{PQ}

2\times PQ=5\times AQ

PQ=\frac{5\:AQ}{2} .............................(1)

In ΔPBQ,

using trigonometric ratio we get

cot\,PBQ=\frac{BQ}{PQ}

\frac{3}{5}=\frac{AB+AQ}{PQ}

3\times PQ=5\times(32+AQ)

3\times\frac{5\:AQ}{2}=5\times(32+AQ)  (from (1))

\frac{15\:AQ}{2}=160+5AQ

\frac{15\:AQ}{2}-5AQ=160

\frac{5\:AQ}{2}=160

AQ=160\times\frac{2}{5}

AQ=64

Now put this value in (1)

PQ=\frac{5\times64}{2}

PQ = 160 m

Therefore, Height of tower is 160 m.

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