The shadow of a tower standing on level ground is found

Solution

Let AB be the tower, BC be the shadow at 45°, and BD be the shadow at 30°. The shadow of a tower standing on a level ground is found to be 40 m longer when the sun's altitude is 30° than when it is 45°. i.e. CD = 40 m Let BC = x m In triangle ABC, we have Tan45 =AB/BC AB =x.       … (1) In triangle ABD, We have. Tan 30 =AB/BD 1/ √ 3 =AB/x+40 √ 3AB =x+40 X =√ 3AB-40 … (2) x = √ 3x-40 x=40/ (√ 3-1)  we multiply the numerator and denominator by the conjugate of the denominator: x=20 (√ 3+1).