Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Let AB be a vertical pole and let its shadow be BC

Let BC = x m, then length of pole = $$\sqrt 3 $$ x,
$$\theta $$ be the angle of elevation
$$\eqalign{ & \therefore {\text{tan}}\theta = \frac{{AB}}{{BC}} = \frac{{\sqrt 3 \,x}}{x} = \sqrt 3 \cr & = \tan {60^ \circ } \cr & \therefore \theta = {60^ \circ } \cr} $$

Let AB and CD are two persons standing ‘a’ meters apart
P is the mid-point of BD and from M, the angles of elevation of A and C are complementary

$$\eqalign{ & {\text{In}}\,\,\Delta {\text{APB,}} \cr & \tan \theta = \frac{{AB}}{{BP}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a} \cr & {\text{In}}\,\,\Delta {\text{CDP,}} \cr & \cot (90 - \theta ) = \frac{{PD}}{{CD}} = \frac{{\frac{a}{2}}}{{2h}} = \frac{a}{{4h}} \cr & {\text{We}}\,\,{\text{Know}}\,\,{\text{that,}} \cr & \tan \theta = \cot (90 - \theta ). \cr & \therefore \frac{{2h}}{a} = \frac{a}{{4h}} \cr & \Rightarrow 8{h^2} = {a^2} \cr & \Rightarrow h = \frac{a}{{2\sqrt 2 }} \cr} $$

Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60°

Let height of the tower = h
$$\eqalign{ & {\text{Then in right }}\Delta ABP \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {60^ \circ } = \frac{h}{{100}} \Rightarrow \sqrt 3 = \frac{h}{{100}} \cr & \Rightarrow h = 100\sqrt 3 \cr & \therefore {\text{Height}}\,{\text{of}}\,{\text{tower}} = 100\sqrt 3 \cr} $$

Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively

∠ACB = ∠XAC = 30° , ∠ADB = ∠YAD = 45° and
CD = 100m
Let AB =h and CB = x then BC = (100 - x)m
$$\eqalign{ & {\text{Now in }}\,\Delta ACB, \cr & \tan \theta = \frac{{AB}}{{CB}} \cr & \tan {30^ \circ } = \frac{h}{x} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr & \Rightarrow x = \sqrt 3 h ......{\text{(i)}} \cr & {\text{Similarly in }}\,\Delta ADB \cr & \tan {45^ \circ } = \frac{{AB}}{{BD}} \cr & \Rightarrow 1 = \frac{h}{{100 - x}} \cr & \Rightarrow x = 100 - h \,......{\text{(ii)}} \cr & {\text{From (i) and (ii)}} \cr & \sqrt 3 h = 100 - h \cr & \Rightarrow (\sqrt 3 - 1)h = 100 \cr & h = \frac{{100}}{{\sqrt 3 + 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{2} \cr & \,\,\,\,\,\, = 50\left( {\sqrt 3 - 1} \right) \cr} $$
∴ height of light house $$ = 50\left( {\sqrt 3 - 1} \right)m$$

AB is a tower and AB = 75 m
From A, the angle of depression of a car C
on the ground is 30°

$$\eqalign{ & {\text{Let distance }}BC = x \cr & {\text{Now in right }}\Delta ACB, \cr & \tan \theta = \frac{{AB}}{{BC}} \cr & \Rightarrow \tan {30^ \circ } = \frac{{75}}{x} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{75}}{x} \cr & \Rightarrow x = 75\sqrt 3 \,m \cr & \therefore BC = 75\sqrt 3 \,m \cr} $$

Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively

$$\eqalign{ & {\text{Let }}AB = h \cr & {\text{Now in right }}\Delta APB, \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {30^ \circ } = \frac{h}{a} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{a}\,...........(i) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan {60^ \circ } = \frac{{AB}}{{QB}} \cr & \Rightarrow \sqrt 3 = \frac{h}{b}\,...........(ii) \cr & {\text{Multiplying (i) and (ii)}} \cr & \frac{1}{{\sqrt 3 }} \times \sqrt 3 = \frac{h}{a} \times \frac{h}{b} \cr & \Rightarrow 1 = \frac{{{h^2}}}{{ab}} \cr & \Rightarrow {h^2} = ab \cr & \Rightarrow h = \sqrt {ab} \cr & \therefore {\text{Height of the tower}} = \sqrt {ab} \cr} $$

Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground

∴ In ∆ABC, ∠B = 90°
Let height of wall AB = h
Then
$$\sin \theta = \frac{{AB}}{{AC}} \Rightarrow \sin {60^ \circ } = \frac{h}{{15}}$$
$$\eqalign{ & \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{h}{{15}} \cr & \Rightarrow h = \frac{{15\sqrt 3 }}{2}\,m \cr} $$
∴ Height of the wall $$ = \frac{{15\sqrt 3 }}{2}\,m$$

Let height of pole CD = h
and AB = 2h, BD = a
M is mid-point of BD

$$\therefore DM = MB = \frac{a}{2}$$
$${\text{Let }}\angle CMD = \theta ,$$    $${\text{then }}\angle AMB = $$     $${90^ \circ } - \theta $$
$$\eqalign{ & {\text{Now}} \cr & \tan \theta = \frac{{CD}}{{DM}} = \frac{h}{{\frac{a}{2}}} = \frac{{2h}}{a}\,......({\text{i}}) \cr & {\text{and}} \cr & tan\left( {{{90}^ \circ } - \theta } \right) = \frac{{AB}}{{MB}} = \frac{{2h}}{{\frac{a}{2}}} = \frac{{4h}}{a} \cr & \Rightarrow \cot \theta = \frac{{4h}}{a}\,..........({\text{ii}}) \cr & {\text{Multiplying (i) and (ii)}} \cr & {\text{tan}}\theta \times {\text{cot}}\theta = \frac{{2h}}{a} \times \frac{{4h}}{a} \cr & 1 = \frac{{8{h^2}}}{{{a^2}}} = {h^2} = \frac{{{a^2}}}{8}\,m \cr & h = \sqrt {\frac{{{a^2}}}{8}} = \frac{a}{{\sqrt 8 }} = \frac{a}{{2\sqrt 2 }}\,m \cr} $$

Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C
Let angle be Q and CD = 25 m

$$\eqalign{ & {\text{Let}}\,AB = h \cr & CE \,\, || \,\, DB \cr & \therefore EC = DB = x{\text{ }}\left( {{\text{suppose}}} \right) \cr & EB = CD = 25 \cr & \therefore AE = h - 25 \cr & {\text{Now in right }}\Delta CDB, \cr & \tan \theta = \frac{{CD}}{{DB}} = \frac{{25}}{x}\,......\left( {\text{i}} \right) \cr & {\text{and in right }}\Delta CAE \cr & \tan \theta = \frac{{AE}}{{CE}} = \frac{{h - 25}}{x}\,......\left( {{\text{ii}}} \right) \cr & {\text{From}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{25}}{x} = \frac{{h - 25}}{x} \cr & \Rightarrow 25 = h - 25 \cr & \Rightarrow h = 25 + 25 = 50 \cr & \therefore {\text{Height of tower}} = 50\,m \cr} $$

Let AB be tower and C is a point on the ground 50 m away

From foot of tower B
Angle of elevation is 45°
Let h be height of tower = x m
$$\eqalign{ & \therefore \tan \theta = \frac{{AB}}{{BC}} \cr & \Rightarrow \tan {45^ \circ } = \frac{{AB}}{50} \cr & \Rightarrow 1 = \frac{{AB}}{{50}} \Rightarrow AB = 50\,m \cr} $$