The shadow of a tower standing on a level plane is found to be 50m longer when sun’s elevation is 300 than when it is 600. Find the height the tower.
Let SQ = h be the tower.
∠SPQ = 30° and ∠SRQ = 60°
According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is 30° than when it was 60°. So,
PR = 50 m and RQ = x m
So in ∆SRQ, we have
In ∆SPQ,
Substituting the value of x in the above equation, we get
⇒ 50√3+h = 3h
⇒ 50√3 = 3h - h
⇒ 3h - h = 50√3
⇒ 2h = 50√3
⇒ h = (50√3)/2
⇒ h = 25√3
Hence, the required height is 25√3 m.