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)?
Answer & explanation
Correct answer: option 1
⇒ 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 \(\sqrt {3 }\)