trigonometry(height and distance)
posted by Anamika .
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.

trigonometry(height and distance) 
Reiny
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 rightangled 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/(√31)
then PQ = √3(5/(√31)) = 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 
trigonometry(height and distance) 
Anamika
Ans is correct can you show me fig.

trigonometry(height and distance) 
Reiny
We can't put diagrams on here, I think you should be able to follow my directions. I described the diagram.
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