The shadow of a tower when the angle of elevation of the sun is 45 degree is found to be 5m longer when it is 60 degree. Find the height of the tower.

Answer 1

How about starting with a diagram.
Label the tower as PQ, with Q on the ground.
Label the point with angle 45° as A, and the point with the 60° angle as B.
You now have two right-angled triangles, PAQ and PBQ

let BQ be x
then tan60 = PQ/x ---> PQ = xtan60 = √3 x
and tan 45 = PQ/(x+5) ---> PQ = (x+5)tan45 = (x+5)(1)

so √3 x =x+5
√3 x - x = 5
x(√3 - 1) = 5
x = 5/(√3-1)
then PQ = √3(5/(√3-1)) = appr 11.83 m

or

look at triangle PAB, angle PBA = 120° making angle APB = 15°
by the sine law:
PB/sin45 = 5/sin15
PB = 5sin45/sin15

in triangle PBQ,
sin60 = PQ/PB
PQ = PBsin60 = (5sin45/sin15)(√3/2) = appr 11.83

In my experience, most students find the second method easier to follow

Answer 2

Ans is correct can you show me fig.

Answer 3

We can't put diagrams on here, I think you should be able to follow my directions. I described the diagram.