Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option B

Let AB be the tower of height 150 m
C is car and angle of depression is 30°

Therefore, ∠ACB = 30° (alternate angle)
In right - angled triangle ABC,
$$\eqalign{ & \frac{{BC}}{{AB}} = \cot {30^ \circ } \cr & \Rightarrow \frac{{BC}}{{150}} = \sqrt 3 \cr & \Rightarrow BC = 150\sqrt 3 \,m \cr} $$
That is, distance of the car from the tower is $$150\sqrt 3 \,m$$

Answer: Option A

Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60°

$$\eqalign{ & {\text{Let height of tower }}AB = h \cr & {\text{Then in right }}\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{{30}} \cr & \Rightarrow \sqrt 3 = \frac{h}{{30}} \cr & \Rightarrow h = 30\sqrt 3 \cr} $$
∴ Height of the tower $$ = 30\sqrt 3 \,m$$

Answer: Option C

Let AB and CD are two poles AB = 10 m and CD = 16 m

AC is wire which makes an angle of 30° with the horizontal
Let BD = x, then AE = x
CE = CD - ED = CD - AB = 16 - 10 = 6m
$$\eqalign{ & {\text{Now}}\,{\text{in}}\,\Delta ACE \cr & \sin {30^ \circ } = \frac{{CE}}{{AC}} = \frac{6}{l} \cr & \Rightarrow \frac{1}{2} = \frac{6}{l} \cr & \Rightarrow l = 2 \times 6 = 12\,m \cr} $$

Answer: Option C

Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD = y

$$\eqalign{ & {\text{Now in right }}\Delta ADB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{DB}} \cr & \tan {45^ \circ } = \frac{{100}}{y} \cr & \Rightarrow 1 = \frac{{100}}{y} \cr & \Rightarrow y = 100 \cr & {\text{Similarly in right }}\Delta ACB, \cr & \tan {30^ \circ } = \frac{{AB}}{{CB}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100}}{{y + x}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100}}{{100 + x}} \cr & \Rightarrow 100 + x = 100\sqrt 3 \cr & \Rightarrow x = 100\sqrt 3 - 100 \cr & \Rightarrow x = 100\left( {\sqrt 3 - 1} \right)\,m \cr} $$

Answer: Option B

Let AB be the tower and C is a point such that the angle of elevation of A is α.
After walking towards the foot B of the tower, at D the angle of elevation is β.
Let h be the height of the tower and DB = x Now in ΔACB,

$$\eqalign{ & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{CB}} \cr & \tan \alpha = \frac{h}{{d + x}} \cr & \Rightarrow d + x = \frac{h}{{\tan \alpha }} \cr & \Rightarrow d + x = h\cot \alpha \cr & \Rightarrow x = h\cot \alpha - d\,.......\left( {\text{i}} \right) \cr & {\text{Similarly in right }}\Delta ADB, \cr & \tan \beta = \frac{h}{x} \cr & \Rightarrow x = \frac{h}{{\tan \beta }} \cr & \Rightarrow x = h\cot \beta \,.........({\text{ii}}) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & h\cot \alpha - d = h\cot \beta \cr & \Rightarrow h\cot \alpha - h\cot \beta = d \cr & \Rightarrow h\left( {\cot \alpha - \cot \beta } \right) = d \cr & \Rightarrow h = \frac{d}{{\cot \alpha - \cot \beta }} \cr} $$
∴ Height of the tower $$ = \frac{d}{{\cot \alpha - \cot \beta }}$$