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
$$\eqalign{ & {\text{Then,}}\frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow AB = AC \cr & \Rightarrow h = 60\,m \cr & {\text{And}}\frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow AD = \left( {AB \times \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 60\sqrt 3 \,m \cr & \therefore CD = \left( {AD - AC} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 60\left( {\sqrt 3 - 1} \right)\,m \cr & {\text{Hence, required speed }} \cr & {\text{ = }}\left[ {\frac{{60\left( {\sqrt 3 - 1} \right)}}{5}} \right]{\text{m/s}} \cr & {\text{ = }}\left( {12 \times 0.73} \right){\text{m/s}} \cr & = \left( {12 \times 0.73 \times \frac{{18}}{5}} \right){\text{km/hr}} \cr & = {\text{31}}{\text{.5km/hr }} \approx {\text{ 32km/hr}} \cr} $$

Answer: Option C

Let AB be the tower and CD be the electric pole.
Then, $$\angle ACB = {60^ \circ },$$     $$\angle EDB = {30^ \circ }$$     and AB = 15 m
Let CD = h
Then, BE = (AB - AE) = (AB - CD) = (15 - h)
$$\eqalign{ & \frac{{AB}}{{AC}} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow AC = \frac{{AB}}{{\sqrt 3 }} = \frac{{15}}{{\sqrt 3 }} \cr & {\text{And, }}\frac{{BE}}{{DE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow DE = \left( {BE \times \sqrt 3 } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 3 \left( {15 - h} \right) \cr & {\text{So, }}AC = DE \cr & \Rightarrow \frac{{15}}{{\sqrt 3 }} = \sqrt 3 \left( {15 - h} \right) \cr & \Rightarrow 3h = \left( {45 - 15} \right) \cr & \Rightarrow h = 10{\text{ m}} \cr} $$

Answer: Option B

$${\text{Given tan }}{x^ \circ } = \frac{2}{5}$$     and AF = 200 meter
$$\eqalign{ & \therefore \frac{2}{5} = \frac{{TF}}{{AF}} \cr & \Rightarrow TF = \frac{{2 \times 200}}{5} \cr & \Rightarrow TF = 80{\text{ m}} \cr & {\text{We have, BF = 80 m}} \cr & \therefore \tan {\text{ }}{y^ \circ } = \frac{{TF}}{{BF}} \cr & \Rightarrow \tan {\text{ }}{y^ \circ } = \frac{{80}}{{80}} \cr & \Rightarrow \tan {\text{ }}{y^ \circ } = 1 = \tan {45^ \circ } \cr & \therefore {y^ \circ } = {45^ \circ } \cr} $$

Answer: Option C

$$\eqalign{ & {\text{Let, }}OP = {\text{ }}a \cr & {\text{tan }}{60^ \circ } = \frac{H}{a} \cr & \Rightarrow H = \sqrt 3 a \cr & \Rightarrow \frac{H}{{\sqrt 3 }} = a.....(i) \cr} $$
$$tan{45^ \circ } = 1$$   $$ = \frac{H}{{a + 100\left( {3 - \sqrt 3 } \right)}}$$
$$ \Rightarrow a + 100\left( {3 - \sqrt 3 } \right) = H$$
From (i) $$\frac{H}{{\sqrt 3 }} + $$   $$100\left( {3 - \sqrt 3 } \right)$$   = H
$$\eqalign{ & \Rightarrow H + 300\sqrt 3 - 300 = \sqrt 3 H \cr & \Rightarrow 300\sqrt 3 - 300 = \sqrt 3 H - H \cr & \Rightarrow \left( {\sqrt 3 - 1} \right)H = 300\left( {\sqrt 3 - 1} \right) \cr & \Rightarrow H = 300{\text{ m}} \cr} $$

Answer: Option A

Let AB be lighthouse and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h
Let PB = x and QB = y

$$\eqalign{ & {\text{Now in right }}\Delta APB \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {45^ \circ } = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \cr & \Rightarrow x = h\,............(i) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan {30^ \circ } = \frac{{AP}}{{QB}} = \frac{h}{y} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{y} \cr & \Rightarrow y = \sqrt 3 \,h\,..............(ii) \cr & {\text{Adding (i) and (ii)}} \cr & \therefore PQ = x + y \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = h + \sqrt 3 \,h \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt 3 + 1} \right)\,h \cr} $$

Answer: Option C

Let AB is girls and CD is lamp-post AB = 1.5
which casts her shadow EB

$$\eqalign{ & \therefore EB = 4.5\,m,\,BD = 3\,m \cr & {\text{Now in }}\Delta AEB \cr & \tan \theta = \frac{{AB}}{{BE}} = \frac{{1.5}}{{4.5}} = \frac{1}{3} \cr & {\text{and in }}\Delta CED \cr & \tan \theta = \frac{{CD}}{{ED}} \Rightarrow \frac{1}{3} = \frac{h}{{4.5 + 3}} \cr & \Rightarrow \frac{1}{3} = \frac{h}{{7.5}} \cr & \Rightarrow h = \frac{{7.5}}{3} - 2.5\,m \cr & \therefore {\text{Height of lamp - post}} = {\text{2}}{\text{.5}}\,m \cr} $$