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+1 vote
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in Trigonometry by (34.5k points)
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The angle of elevation of a top of the tower from a point 120 m. from its foot is 45°. How much the tower is to be raised when the elevation is to be 60° at the same point ?

A) 120 √3 m 

B) 120 m 

C) 120(√3 + 1) m 

D) 120(√3 – 1) m

2 Answers

+1 vote
by (66.4k points)
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Best answer

Correct option is: D) 120(√3 – 1) m

Let  \(\angle\)CAB =\(45^\circ\) &  \(\angle\)DAB = \(60^\circ\)

Also let BC be the original height of the tower and CD be the length to be raised to make angle of elevation \(60^\circ\) at point A.

In right  \(\triangle\)ABC,

tan A = \(\frac {BC}{AB}\)

= BC = AB tan A = 120 tan \(45^\circ\) (\(\because\) \(\angle\)CAB =\(45^\circ\))

= BC =  120 m ....(1) (\(\because\) tan \(45^\circ\) = 1)

In right  \(\triangle\)ABC,

tan A = \(\frac {BD}{AB}\)

= BD = AB tan A = AB tan \(60^\circ\) (\(\because\)  \(\angle\)DAB \(60^\circ\))

= BD = 120\(\sqrt3\) ...(2) (\(\because\) AB = 120 m)

Now, CD = BD-BC

= 120\(\sqrt3\) -120 (From (1) & (2)

= 120 (\(\sqrt3\)-1)m

Hence, tower to be raised 120 (\(\sqrt3\)-1)m to make angle of elevation \(60^\circ\) at the same point.

+2 votes
by (35.6k points)

Correct option is: D) 120(√3 – 1) m

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