Question

the shadow of a tower at s same time of the day is 3 times as long as shadow of the tower . when the cycle of elevation of the sun is 60. find the angle of elevation of the sun at the time of longer shadow

Answer 1

given BD= 3 BC...
In triangle ABC,
tan =  60° AB/ BC 
∴ √ 3= h/BC
⇒ h = √ 3BC
In triangle ABC, 
tanθ= AB/BD
From 1,2,3 we get
√ 3BC tanθ× 3BC
∴ tan θ = √ 3/3 = 1/√ 3
θ=30° (∵ tan 30 = 1/√3)
∴ answer is 30°

Answer 2

ans=30°
let the tower be AB the small shadow be BC(x) and long one be BD
atq
BD=3BC=3x
In ABC
AB/BC=AB/X=tan 60=√3
AB=x√3          (1)
In ABD
AB/BD=AB/3x=tan Y
FROM (1)
x√3/3x=tan Y
1/√3=tan Y
∴∠Y(∠ADB)=30°