Question

Two men on either side of a temple 75m high boserve angle of elevation of top 75m hugh to be 30deg and 60deg. Find the distabce between the two men

Answer 1

It forms two triangles here in this case.Hence opposite sides of the given angles are given from which we can calculate adjacent distance between two persons.

From 1st triangle we get tan 60=75/x

from this we get x=15√3

From 2nd triangle we get tan 30=75/y.From this we can calculate y= 75√3

Hence distance between them is x+y=75√3+15√3=90√3

Answer 2

Let AB be the height of temple =75m

P and Q be the positions of the two men on either side of the temple.

Then,

$\sf{\angle{APB}=30°}$

And $\sf{\angle{AQB}=60°}$

From right ∆ABP,we have

$tex\sf{tan30° = \frac{AB}{PB}}$

=$\sf{\frac{1}{ \sqrt{3} } = \frac{75}{PB}}$

= $\sf{PB = 75 \sqrt{3} m}$

From right ∆AQB,we have

$\sf{tan60°=\frac{AB}{BQ}}$

=$\sf{√3=\frac{75}{BQ}}$

=$\sf{BQ=\frac{75}{√3}}$

=$\sf{25√3m}$

Hence, the distance between the two men is

$\longrightarrow$$\sf{PQ=PB+BQ}$

$\longrightarrow$$\sf{75√3+25√3}$

$\longrightarrow$${\boxed{\sf{100√3m}}}$