Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 200 m, ∠BAD = 30°, ∠BCD = 45°
$$\eqalign{ & \tan {30^ \circ } = \frac{{BD}}{{BA}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{200}}{{BA}} \cr & \Rightarrow BA = 200\sqrt 3 \cr & \tan {45^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow 1 = \frac{{200}}{{BC}} \cr & \Rightarrow BC = 200 \cr} $$
Distance between the two ships
$$\eqalign{ & = AC = BA + BC \cr & = 200\sqrt 3 + 200 \cr & = 200(\sqrt 3 + 1) \cr & = 200\left( {1.73 + 1} \right) \cr & = 200 \times 2.73 \cr & = 546{\text{ }}m \cr} $$

Answer: Option D

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have ∠DAC = 30°, ∠DBC = 45°
Let DC = h
$$\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{AC}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{AC}} \cr & \Rightarrow AC = h\sqrt 3 \,.........\,\left( {\text{i}} \right) \cr & \tan {45^ \circ } = \frac{{DC}}{{BC}} \cr & \Rightarrow 1 = \frac{h}{{BC}} \cr & \Rightarrow BC = h\,........\,\left( {{\text{ii}}} \right) \cr & {\text{We know that,}} \cr & AB = \left( {AC - BC} \right) \cr & \Rightarrow 40 = \left( {AC - BC} \right) \cr} $$
$$ \Rightarrow 40 = \left( {h\sqrt 3 - h} \right)$$   [∵ from (1) and (ii)]
$$\eqalign{ & \Rightarrow 40 = h\left( {\sqrt 3 - 1} \right) \cr & \Rightarrow h = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \cr & = \frac{{40}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{{\left( {3 - 1} \right)}} \cr & = \frac{{40\left( {\sqrt 3 + 1} \right)}}{2} \cr & = 20\left( {\sqrt 3 + 1} \right) \cr & = 20\left( {1.73 + 1} \right) \cr & = 20 \times 2.73 \cr & = 54.6\,m \cr} $$

Answer: Option D

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.
$$\eqalign{ & \cos {60^ \circ } = \frac{{PQ}}{{PR}} \cr & \frac{1}{2} = \frac{{12.4}}{{PR}} \cr & PR = 2 \times 12.4 \cr & \,\,\,\,\,\,\,\,\,\, = 24.8\,m \cr} $$

Answer: Option B

Let RQ be the pole and PQ be the shadow
Given that RQ = 18 m and PQ = $$6\sqrt 3 $$ m
Let the angle of elevation, ∠RPQ = $$\theta $$
$$\eqalign{ & {\text{From the right}}\Delta PQR, \cr & {\text{tan}}\,\theta = \frac{{RQ}}{{PQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{6\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \cr & \Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) = {60^ \circ } \cr} $$

Answer: Option B

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.
Given that CA = 150 m, ∠BCA = 60°
$$\eqalign{ & \tan {60^ \circ } = \frac{{BA}}{{CA}} \cr & \sqrt 3 = \frac{{BA}}{{150}} \cr & BA = 150\sqrt 3 \cr} $$
i.e, the distance travelled by the balloon = $$150\sqrt 3 $$   meters
time taken = 2 min = 2 × 60 = 120 seconds
$$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{150\sqrt 3 }}{{120}} = 1.25\sqrt 3 \cr} $$
= 1.25 × 1.73 = 2.16 meter/second