Question

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)

Answer 1

hope this is helpful.......

Answer 2

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.