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Q.

The shadow of a tower, when the angle of elevation of the sun is $45^{\circ}$ , is found to be 10 metres longer than when the angle of elevation is $60^{\circ}$ . Find the height of the tower.

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a

15 + 5 \sqrt{3} m

b

12 + 5 \sqrt{3} m

c

15 + \sqrt{3} m

d

12 + \sqrt{3} m

answer is A.

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Detailed Solution

Suppose, Tower height = AB
AC = shadow length at angle of elevation are

60^{\circ}

AD = shadow length at angle of elevation are

45^{o}

Question Image

In right

△ CAB

,

\tan 60^{\circ} = \frac{AB}{AC}

\sqrt{3} = \frac{AB}{AC}

AC = \frac{AB}{\sqrt{3}}

…………… (i)
In right

△ DAB

,

\tan 45^{\circ} = \frac{AB}{AD}

1 = \frac{AB}{10 + AC}

AC = AB - 10

Put value of AC from (i),

\frac{AB}{\sqrt{3}} = AB - 10

AB = \sqrt{3} AB - 10 \sqrt{3}

\sqrt{3} AB - AB = 10 \sqrt 3

(\sqrt{3} - 1) AB = 10 \sqrt{3}

AB = \frac{10 \sqrt{3}}{(\sqrt{3} - 1)}

Rationalize the denominator,

AB = \frac{[10 \sqrt{3} \times (\sqrt{3} + 1)]}{3 - 1}

AB = 5 (3 + \sqrt{3})

AB = 15 + 5 \sqrt{3} m

This is the required height of tower.
Correct option is 1

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