Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option B

Let AB be tower and BC be its shadow
∴ Let AB = x

$$\eqalign{ & {\text{Then}}\,BC = \sqrt 3 \times x = \sqrt 3 \,x \cr & \therefore \tan \theta = \frac{{AB}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{x}{{\sqrt 3 \,x}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \tan {30^ \circ } \cr & \therefore \theta = {30^ \circ } \cr} $$
∴ Angle of elevation of the sun$${\text{ = }}{30^ \circ }$$

Answer: Option A

AB is a tower
BD and BC are its shadows and CD = 2x

$$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{DB}} \cr & \Rightarrow 1 = \frac{h}{y} \Rightarrow y = h \cr & {\text{and}}\tan {30^ \circ } = \frac{{AB}}{{CB}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{2x + y}} \cr & \Rightarrow 2x + y = \sqrt 3 \,h \cr & \Rightarrow \sqrt 3 \,h - h = 2x \cr & \Rightarrow h\left( {\sqrt 3 - 1} \right) = 2x \cr & \Rightarrow h = \frac{{2x}}{{\sqrt 3 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {\sqrt 3 + 1} \right) \cr} $$

Answer: Option B

Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are $$\theta $$ and (90° – $$\theta $$)
Let height of tower = h

$$\eqalign{ & {\text{Then}}\,{\text{in}}\,{\text{right}}\,\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & = \frac{h}{a}\,............\left( {\text{i}} \right) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan \left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{QB}} = \frac{h}{b} \cr & \Rightarrow \cot \theta = \frac{h}{b}\,...........\left( {{\text{ii}}} \right) \cr & {\text{Multiplying }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \tan \theta \cot \theta = \frac{h}{a} \times \frac{h}{b} \cr & \Rightarrow 1 = \frac{{{h^2}}}{{ab}} \cr & \Rightarrow {h^2} = ab \cr & \Rightarrow h = \sqrt {ab} \cr & \therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = \sqrt {ab} \cr} $$

Answer: Option D

Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m
∠BAD = 45°, ∠BCD = 60°
$$\eqalign{ & \tan {45^ \circ } = \frac{{BD}}{{BA}} \cr & \Rightarrow 1 = \frac{{60}}{{BA}} \cr & \Rightarrow BA = 60\,m\,........\left( {\text{i}} \right) \cr & \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{{60}}{{BC}} \cr & \Rightarrow BC = \frac{{60}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{60\sqrt 3 }}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 34.6\,m\,..........\left( {{\text{ii}}} \right) \cr} $$
Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6 [∵ Substituted value of BA and BC from (i) and (ii)]
= 94.6 m

Answer: Option D

Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m, ∠DAC = 45°, ∠DBC = 60°
$$\eqalign{ & \tan {60^ \circ } = \frac{{DC}}{{BC}} \cr & \sqrt 3 = \frac{{600}}{{BC}} \cr & BC = \frac{{600}}{{\sqrt 3 }}\,..........\,\left( {\text{i}} \right) \cr & \tan {45^ \circ } = \frac{{DC}}{{AC}} \cr & 1 = \frac{{600}}{{AC}} \cr & AC = 600\,............\,\left( {{\text{ii}}} \right) \cr} $$
Distance between the objects
$$ = AC = \left( {AC - BC} \right)$$
$$ = 600 - \frac{{600}}{{\sqrt 3 }}$$   [∵ from (1) and (ii)]
$$\eqalign{ & = 600\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{600\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{3} \cr & = 200\sqrt 3 \left( {\sqrt 3 - 1} \right) \cr & = 200\left( {3 - \sqrt 3 } \right) \cr & = 200\left( {3 - 1.73} \right) \cr & = 254\,m \cr} $$

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