Question

The angle of elevation of a tower from a point on the ground is 30degree. at a point on horizontal line passing through the foot of the tower and 100 metres nearer it, the angle of elevation is found to be 60degree. find the height of the tower and the distance of the first point from the tower

Answer 1

"Angle of elevation of top of the tower from point A, α=30∘.

Angle of elevation of top of tower from point B, β=60∘.

Distance between A and B, AB = 100 m

Let height of tower CD = ‘h’ m

Distance between second point B from foot of the tower be ‘x’ m

If we represent the above data in form of figure then it form a figure as shown with ∠D=90∘

In right angle triangle, one of included angle is Θ

Then tanΘ=oppositeside/adjacentside

tanα=CD/AD

tan30∘=h/100+x

100 + x = h3–√ ---- (a)

tanβ=CD/BD

tan60∘=h/x

x = h/√3 ---- (b)

Substituting (b) in (a)

100 + h/√3=h√3

h(√3−1/√3)=100

h(3−1/√3)=100

h=100×√3/2

h = 50√3

h = 86.60 m

x = 50√3/√3

x = 50 m

Height of the tower = 17.32 m

Distance of the tower from point A = (100 + 50) = 150 m

"