Question : On the ground, there is a vertical tower with a flagpole on its top. At a point 9 metres away from the foot of the tower, the angles of elevation of the top and bottom of the flagpole are 60° and 30°, respectively. The height of the flagpole is:
Option 1: $5\sqrt{3}$ metres
Option 2: $6\sqrt{3}$ metres
Option 3: $6\sqrt{2}$ metres
Option 4: $6\sqrt{5}$ metres
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Correct Answer: $6\sqrt{3}$ metres
Solution : In the figure, the flagpole is AB, BC is the tower and CD is the base. In $\triangle$ACD, $⇒\tan 60°=\frac{\text{AC}}{\text{DC}}$ $⇒\sqrt{3}=\frac{\text{AC}}{9}$ $⇒\text{AC}=9\sqrt{3}$ metres In $\triangle$BCD, $⇒\tan 30°=\frac{\text{BC}}{\text{DC}}$ $⇒\frac{1}{\sqrt{3}}=\frac{\text{BC}}{9}$ $⇒\text{BC}=\frac{9}{\sqrt{3}} = 3\sqrt{3}$ metres $⇒\text{AC = AB + BC}$ $⇒\text{AB = AC – BC}$ $⇒\text{AB} = 9\sqrt{3} - 3\sqrt{3}=6\sqrt{3}$ metres Hence, the correct answer is $6\sqrt{3}$ metres.
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Question : Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 45°, respectively. If the height of the tower is 50 metres, the distance between the two men is: (take $\sqrt{3}=1.732$)
Question : From point P on the level ground, the angle of elevation to the top of the tower is 30°. If the tower is 100 metres high, the distance of point P from the foot of the tower is: (Take $\sqrt{3}$ = 1.73)
Question : From the top of a 20 metres high building, the angle of elevation from the top of a tower is 60° and the angle of depression of its foot is at 45°, then the height of the tower is: $(\sqrt{3} = 1.732)$
Question : The shadow of a tower when the angle of elevation of the sun is 45°, is found to be 10 metres longer than when it was 60°. The height of the tower is:
Question : From the top of an upright pole 17.75 m high, the angle of elevation of the top of an upright tower was 60°. If the tower was 57.75 m tall, how far away (in m) from the foot of the pole was the foot of the tower?
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