Question

Find the angle of elevation of the sun’s rays from a point on the ground at a distance of 3√3m, from the foot of tower 3m high.

Answer 1

Let AC be the height of the pole AC = 3mts & AB = 3√3m mts 

From the triangle ABC we have 

tan θ = \(\frac{AC}{AB}\) = \(\frac{3}{3\sqrt3}\) = \(\frac{1}{\sqrt3}\) = tan 30° 

∴ The angle of elevation is 30°