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The angle of depression of a point situated at a distance of 70m from the base of a tower is 60°. The height of the tower is-

a) \[35\sqrt 3 {\text{ m}}\]
b) \[70\sqrt 3 {\text{ m}}\]
c) \[\frac{{70\sqrt 3 }}{3}{\text{ m}}\]
d) \[{\text{70 m}}\]

Answer: b
Explanation:
q49
Length of the tower AB = h meter.
$$\eqalign{ & \angle DAC = \angle ACB = {60^ \circ } \cr & BC = 70{\text{ meter}} \cr & {\text{In }}\vartriangle {\text{ABC,}} \cr & {\text{tan }}{60^ \circ } = \frac{{AB}}{{BC}} \cr & \sqrt 3 = \frac{h}{{70}} \cr & h = 70\sqrt 3 {\text{ meter}} \cr} $$

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