Question

Two men standing on either side of a tower is 60 metre high observe the angles of elevation of the top of the tower be 45 degree and 60 degree respectively find the distance between the two men.​

Answer 1

★ Refer to the attachment for diagram ★

In Δ ABC

$: \mapsto \tt \tan(45) = \frac{60}{BC}$

$: \mapsto \tt BC = \frac{60}{1}$

$: \mapsto \tt BC = 60 \: \: cm$

And , in Δ ACD

$: \mapsto \tt \tan(60) = \frac{60}{CD}$

$: \mapsto \tt CD = \frac{60}{ \sqrt{3} }$

$: \mapsto \tt CD = 20 \sqrt{3} \: \: cm$

Now , the distance between the two men will be

$: \implies \tt BC + CD$

$: \implies \tt 60 + 20 \sqrt{3} </p><p>$

$: \implies \tt 20(3 + \sqrt{3} )</p><p>$

$: \implies \tt 20 \times 4.732$

$: \implies \tt 94.64 \: \: m$

★ The distance b/w the two men is 94.64 m