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A man of height $6$ ft observed the top of a tower at the top and the foot of a pole of height $10 ft$ are $45^{o}$ and $30^{o}$ of elevation and depression respectively. The height of the tower is?

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13.79 ft

14.59 ft

14.29 ft

none of these

answer is D.

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Detailed Solution

It is given that a man of height

6 ft

observed the top of a tower at the top and the foot of a pole of height

10 ft

are

45^{o}

and

30^{o}

of elevation and depression respectively.
Let AB be the man and CD be the lower then

∠ DAE = 45^{o}

and

∠ ACB = 30^{o}

△ ABC,

\tan 30^{o} = \frac{AB}{BC}

⇒

\frac{1}{\sqrt{3}} = \frac{6}{BC \Rightarrow BC = 6 \sqrt{3} ft}

BC = AE =

6 \sqrt{3} f

t
In triangle ADE,

\tan 4 5^{\circ} = \frac{DE}{AE}

⇒

1 = \frac{DE}{6 \sqrt{3}} .

⇒

DE = 6 \sqrt{3} ft

Height of tower,

CD = CE + DE = AB + DE

⇒ CD

= 6 + 6 \sqrt{3}

⇒

CD = 6 + 6 \times 1.732

⇒ CD

= 6 + 10.392

⇒ CD

= 16.391 ft

Hence, the correct option is 4.

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A man of height 6ft observed the top of a tower at the top and the foot of a pole of height 10 ft are 45oand 30oof elevation and depression respectively. The height of the tower is?

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A man of height $6$ ft observed the top of a tower at the top and the foot of a pole of height $10 ft$ are $45^{o}$ and $30^{o}$ of elevation and depression respectively. The height of the tower is?