The angular elevation of a tower from a point is 30∘. As we more 100m nearer to the base of the tower the angle of elevation becomes 60∘. What is the height of the tower and the distance of the first point from the base of the tower?

Given that,PQ=100mThe angles of elevation at P and Q = 30∘and 60∘AB is representing the tower,P,Q are the points of observation,We have the formula,tanθ=heightbase Considering the required height of the tower AB=h metres and BQ=x metres.We have to find the height,In ΔBAQ,hx=tan60∘ ⇒h=x tan60∘⇒h=x3..........(i) In ΔABP,h100+x=tan30∘ ⇒h100+x=13⇒h3=100+x ⇒x3×3=100+x⇒3x=100+x⇒2x=100⇒x=50 From (i) we get,h=503m ⇒h=50×1.732m ⇒h=86.6m Then, the distance of the first point from the base will be,AP=(x+100)m ⇒AP= (50+100)m ⇒AP= 150m Thus, the height of the tower is 86.6m and the distance is 150m.Therefore the correct option is 1.

The angular elevation of a tower from a point is 30∘. As we more 100m nearer to the base of the tower the angle of elevation becomes 60∘. What is the height of the tower and the distance of the first point from the base of the tower?

Given that,PQ=100mThe angles of elevation at P and Q = 30∘and 60∘AB is representing the tower,P,Q are the points of observation,We have the formula,tanθ=heightbase Considering the required height of the tower AB=h metres and BQ=x metres.We have to find the height,In ΔBAQ,hx=tan60∘ ⇒h=x tan60∘⇒h=x3..........(i) In ΔABP,h100+x=tan30∘ ⇒h100+x=13⇒h3=100+x ⇒x3×3=100+x⇒3x=100+x⇒2x=100⇒x=50 From (i) we get,h=503m ⇒h=50×1.732m ⇒h=86.6m Then, the distance of the first point from the base will be,AP=(x+100)m ⇒AP= (50+100)m ⇒AP= 150m Thus, the height of the tower is 86.6m and the distance is 150m.Therefore the correct option is 1.