A and B are two points on the same side of a ground, 50 metres apart. The angles of elevation of these points to the top of a tree are 60° and 30°, respectively. What is 40% of the height of the tree (in m)?

$10\sqrt{3}$

⇒ tan \({60}^\circ\) = \(\frac{T}{D}\)

⇒ \(\sqrt {3 }\) = \(\frac{T}{D}\)

⇒ T = \(\sqrt {3 }\)D    ..(1)

And,

⇒ tan \({30}^\circ\) = \(\frac{T}{D + 50}\)

⇒ \(\frac{1}{√3}\) = \(\frac{T}{D + 50}\)

⇒ T = \(\frac{D + 50}{√3}\)     ..(2)

Now,

⇒ \(\sqrt {3 }\)D = \(\frac{D + 50}{√3}\)

⇒ 3D = D + 50

⇒ 2D = 50

⇒ D = 25,

Now,

Putting value of D in equation, 1.

⇒ \(\sqrt {3 }\)D =\(\sqrt {3 }\)25

⇒ 40% of \(\sqrt {3 }\)25 = 10 \(\sqrt {3 }\)