Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option B

$$\eqalign{ & {\text{AC = 30 meter}} \cr & {\text{AB = 15 meter}} \cr & \angle {\text{ACB}} = \theta \cr & \therefore \sin \theta = \frac{{AB}}{{AC}} = \frac{{15}}{{30}} = \frac{1}{2} \cr & \Rightarrow \sin \theta = \sin {30^ \circ } \cr & \Rightarrow \theta = {30^ \circ } \cr} $$

Answer: Option D

$$\eqalign{ & {\text{tan }}\theta = \frac{H}{{{m^2}}} \cr & \Rightarrow \tan \left( {90^ \circ - \theta } \right) = \frac{H}{{{n^2}}} \cr & \Rightarrow \cot \theta = \frac{H}{{{n^2}}} \cr & \Rightarrow \tan \theta .\cot \theta = \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} \cr & \Rightarrow \frac{H}{{{m^2}}} \times \frac{H}{{{n^2}}} = 1 \cr & \Rightarrow {H^2} = {m^2}{n^2} \cr & \Rightarrow H = mn \cr} $$

Answer: Option A

$$\eqalign{ & {\text{Here, AD is parallel to BC and AC is a transversal}}{\text{.}} \cr & \Rightarrow {\theta _1} = {\theta _2} \cr} $$

Answer: Option A

$$\eqalign{ & {\text{Let, }}AC = H{\text{ }}m \cr & OA = {\text{ }}a{\text{ }}m \cr & \tan {\text{ }}{30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{H}{a} \cr} $$
$$\tan {60^ \circ } = \sqrt 3 $$    $$ = \frac{{H + 200 + 200}}{a}$$
$$\eqalign{ & \Rightarrow \frac{{H + 400}}{{\sqrt 3 H}} = \sqrt 3 \cr & \Rightarrow H + 400 = 3H \cr & \Rightarrow 2H = 400\,m \cr & \Rightarrow H = 200\,m \cr} $$
Height of cloud above lake = 200 m

Answer: Option A

Let AB be the tower and C and D be the two positions of the boats.
Then, $$\angle {\text{ACB = }}{45^ \circ },$$    $$\angle ADB = {30^ \circ }$$     and AC = 60 m
Let, AB = h
$$\eqalign{ & {\text{Then,}}\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow AB = AC \cr & \Rightarrow h = 60\,m \cr & {\text{And}}\frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow AD = \left( {AB \times \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 60\sqrt 3 \,m \cr & \therefore CD = \left( {AD - AC} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 60\left( {\sqrt 3 - 1} \right)\,m \cr & {\text{Hence, required speed }} \cr & {\text{ = }}\left[ {\frac{{60\left( {\sqrt 3 - 1} \right)}}{5}} \right]{\text{m/s}} \cr & {\text{ = }}\left( {12 \times 0.73} \right){\text{m/s}} \cr & = \left( {12 \times 0.73 \times \frac{{18}}{5}} \right){\text{km/hr}} \cr & = {\text{31}}{\text{.5km/hr }} \approx {\text{ 32km/hr}} \cr} $$