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RUPESH KUMAR

· started a discussion

· 1 Months ago

Wrtite 18 (Question: 111625.0 )

30 mitre ki jagah 18 metre hoga sir

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

Knowledge Expert

· commented

· 1 Months ago

Dear Student,
We rectified it.
You can check it now.

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