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AB is a vertical pole with $B$ at the ground level and $A$ at the top. A man finds that the angle of elevation of the point A from a certain point $C$ on the ground is $60^{o}$ . He moves away from the pole along the line $BC$ to a point $D$ such that $CD = 7$ m. From $D$ the angle of elevation of the point $A$ is $45^{o}$ . Determine the height of the pole.

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\frac{7 \sqrt{3}}{2}

(\sqrt{3} - 1) m

\frac{7 \sqrt{3}}{2} (\frac{1}{\sqrt{3} + 1}) m

\frac{7 \sqrt{3}}{2} (\frac{1}{\sqrt{3} - 1}) m

\frac{7 \sqrt{3}}{2} (\sqrt{3} + 1) m

answer is D.

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Detailed Solution

It is given that angle of elevation of point C and D on the ground to the top of a vertical pole AB is

60

and

45

° respectively.
Let h be the height of the pole AB.
Question Image

From the figure, we have

\tan 45^{0} = \frac{h}{7 + BC}

⇒

1 = \frac{h}{7 + BC}

⇒

7 + BC = h

⇒

BC = h - 7

…(i)
From the figure, we have,

\tan 60^{0} = \frac{h}{BC}

From (i),
⇒

\sqrt{3} = \frac{h}{h - 7}

⇒

\sqrt{3} h - 7 \sqrt{3} = h

⇒

\sqrt{3} h - h = 7 \sqrt{3}

⇒

h = \frac{7 \sqrt{3}}{\sqrt{3} - 1} m

⇒

h = \frac{7 \sqrt{3} (\sqrt{3} + 1)}{2}

Hence, the correct option is 4.

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