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· 1 Months ago

please improve errors

Question:

The angle of elevation of the top of a tree of height 18 m is 30° when measured from a point P in the plane of its base. The distance of the base of the tree from P is:

Options:
A) 6m
B) 6\(\sqrt[]{3}\) m
C) 18 m
D) 18\(\sqrt[]{3}\) m
Solution:
Ans: (d)


Let AB = 18 m be the tree and \(\angle \)APB = 30°

from \(\triangle\)APB,

tan 30o = \(\cfrac{AB}{PB}\)

or \(\cfrac{1}{\sqrt{3}}=\cfrac{18}{PB}\)

or PB = \(18\sqrt{3}\) m

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