The angle of elevation of the height point of a temple situated East of him is 60 degrees on walking 240m to north the angle of elevation is reduced to 30 degrees then the height of the temple is?

This would be easier if I could draw a diagram.

Let the height of the temple be x.

Let the distance between the base of the temple and the place where he measures the 60° from be x

Then h/x=tan60° ==> h = x*sqrt(3) (I use * for multiply) Eqn 1

Also h/(x+240)=tan30° ==> h = (x+240)/sqrt(3) Eqn 2

So x*sqrt(3) = (x+240)/sqrt(3) (as both are equal to h form Eqn 1 and Eqn 2)

==> x*sqrt(3)*sqrt(3) = x + 240 (multiply each side by sqrt(3)

==>3x =x + 240

==> 2x = 240

==> x = 120

Now using Eqn 1, h = x*sqrt(3) = 120 * sqrt3 = 207.8 (to 1decimal place)

So the temple is 207.8m high.