Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option A

Let AB be rod and BC be its shadow
So that AB : BC = 1 : $$\sqrt 3 $$
Let $$\theta $$ be the angle of elevation

$$\eqalign{ & \therefore \tan \theta = \frac{{AB}}{{BC}} = \frac{1}{{\sqrt 3 }} = \tan {30^ \circ } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \tan {{30}^ \circ } = \frac{1}{{\sqrt 3 }}} \right) \cr & \therefore \theta = {30^ \circ } \cr} $$
∴ Hence angle of elevation $$ = {30^ \circ }$$

Answer: Option C

In the figure, AB is chimney and CB and DB are its shadow

$$\eqalign{ & \tan {60^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{h}{{BC}} \cr & \Rightarrow BC = \frac{h}{{\sqrt 3 }}\,.......\,\left( {\text{i}} \right) \cr & {\text{and}} \cr & \tan {30^ \circ } = \frac{h}{{DB}} = \frac{h}{{DB + BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{{x + BC}} \cr & x + BC = h\sqrt 3 \cr & \Rightarrow BC = h\sqrt 3 - x\,.......\,\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & \frac{h}{{\sqrt 3 }} = h\sqrt 3 - x \cr & \Rightarrow \frac{h}{{\sqrt 3 }} - h\sqrt 3 = - x \cr & x = h\sqrt 3 - \frac{h}{{\sqrt 3 }} \cr & x = h\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) \cr & x = h\frac{{3 - 1}}{{\sqrt 3 }} \cr & x = \frac{{2h}}{{\sqrt 3 }} \cr & \therefore h = \frac{{\sqrt 3 }}{2}x \cr} $$

Answer: Option D

Suppose AB is the ladder of length x m
∴ OA = 2m, ∠OAB = 60°

$$\eqalign{ & {\text{In right }}\Delta AOB,\,\sec {60^ \circ } = \frac{x}{2} \cr & \Rightarrow 2 = \frac{x}{2} \cr & \Rightarrow x = 4\,m \cr} $$

Answer: Option A

Let AB and CD be two poles
AB = 20 m, CD = 14 m
A and C are joined by a wire
CE || DB and angle of elevation of A is 30°
Let CE = DB = x and AC = L

Now AE = AB - EB = AB - CD = 20 - 14 = 6 m
$$\eqalign{ & {\text{Now in right }}\Delta ACE, \cr & \sin \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}} = \frac{{AE}}{{AC}} \cr & \Rightarrow \sin {30^ \circ } = \frac{6}{{AC}} \cr & \Rightarrow \frac{1}{2} = \frac{6}{{AC}} \cr & \Rightarrow AC = 2 \times 6 = 12 \cr & \therefore {\text{Length of AC}} = 12\,m \cr} $$

Answer: Option C

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°

$$\eqalign{ & {\text{The}}\,AB = h\,m \cr & {\text{Let}}\,CD = H\,m\,\,\,{\text{and }}\,AD\, = x \cr & {\text{Now}}\,{\text{in}}\,{\text{right}}\,\Delta ABD, \cr & \tan \theta = \frac{{AB}}{{AD}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{x} \cr & \Rightarrow \sqrt 3 = \frac{h}{x} \cr & \Rightarrow x = \frac{h}{{\sqrt 3 }}\,.......({\text{i}}) \cr & {\text{Similarly in right }}\Delta ACB, \cr & \tan {30^ \circ } = \frac{{CD}}{{AD}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{H}{x} \cr & \Rightarrow x = \sqrt 3 \,H\,.......\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & \sqrt 3 \,H = \frac{h}{{\sqrt 3 }} \cr & H = \frac{h}{{\sqrt 3 \times \sqrt 3 }} = \frac{h}{3} \cr & \therefore {\text{Height of tower}} = \frac{h}{3} \cr} $$