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A lower subtends an angle of 30° at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is 60°. The height of the tower is

A. $$\frac{h}{2}\,m$$

B. $$\sqrt 3 \,h\,m$$

C. $$\frac{h}{3}\,m$$

D. $$\frac{h}{{\sqrt 3 }}\,m$$

Answer: Option C

Solution(By Examveda Team)

Let CD is the tower and A is a point such that the angle of elevation of C is 30°
B is and their point h m high of A and angle of depression of D is 60°
Height and Distance mcq solution image
$$\eqalign{ & {\text{The}}\,AB = h\,m \cr & {\text{Let}}\,CD = H\,m\,\,\,{\text{and }}\,AD\, = x \cr & {\text{Now}}\,{\text{in}}\,{\text{right}}\,\Delta ABD, \cr & \tan \theta = \frac{{AB}}{{AD}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{x} \cr & \Rightarrow \sqrt 3 = \frac{h}{x} \cr & \Rightarrow x = \frac{h}{{\sqrt 3 }}\,.......({\text{i}}) \cr & {\text{Similarly in right }}\Delta ACB, \cr & \tan {30^ \circ } = \frac{{CD}}{{AD}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{H}{x} \cr & \Rightarrow x = \sqrt 3 \,H\,.......\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & \sqrt 3 \,H = \frac{h}{{\sqrt 3 }} \cr & H = \frac{h}{{\sqrt 3 \times \sqrt 3 }} = \frac{h}{3} \cr & \therefore {\text{Height of tower}} = \frac{h}{3} \cr} $$

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