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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.

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