Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Let AD be the flagstaff and CD be the building.
Assume that the flagstaff and building subtend equal angles at point B.
Given that AD = 50 m, CD = h and BC = 200 m
Let ∠ABD = $$\theta $$, ∠DBC = $$\theta $$   (∵ flagstaff and building subtend equal angles at a point on level ground).
Then, ∠ABC = 2$$\theta $$
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta BCD, \cr & \tan \theta = \frac{{DC}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{200}}\,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta BCA, \cr & \tan 2\theta = \frac{{AC}}{{BC}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AD + DC}}{{200}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 + h}}{{200}} \cr} $$
$$ \Rightarrow \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \frac{{50 + h}}{{200}}$$      $$\left( {\because \tan \left( {2\theta } \right) = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)$$
$$\frac{{2\left( {\frac{h}{{200}}} \right)}}{{1 - \frac{{{h^2}}}{{{{200}^2}}}}} = \frac{{50 + h}}{{200}}$$      (∵ substituted value of tan $$\theta $$ from eq:1)
$$\eqalign{ & \Rightarrow 2h = \left( {1 - \frac{{{h^2}}}{{{{200}^2}}}} \right)\,\left( {50 + h} \right) \cr & \Rightarrow 2h = 50 + h - \frac{{50{h^2}}}{{{{200}^2}}} - \frac{{{h^3}}}{{{{200}^2}}} \cr} $$
$$ \Rightarrow 2\left( {{{200}^2}} \right)h = 50{\left( {200} \right)^2} + $$     $$h{\left( {200} \right)^2} - $$   $$50{h^2} - {h^3}$$
(∵ multiplied LHS and RHS by $${{{200}^2}}$$ )

h³ + 50h² + (200)²h - (200)²50 = 0

Consider the diagram shown above. AC represents the hill and DE represents the pole
Given that AC = 100 m
∠XAD = ∠ADB = 30° (∵ AX || BD )
∠XAE = ∠AEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h,
AB = (100 - h)   (∵ AC = 100 and BC = h),
BD = CE
$$\eqalign{ & \tan {60^ \circ } = \frac{{AC}}{{CE}} \cr & \Rightarrow \sqrt 3 = \frac{{100}}{{CE}} \cr & \Rightarrow CE = \frac{{100}}{{\sqrt 3 }}\,......\left( 1 \right) \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100 - h}}{{BD}} \cr} $$
$$ \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100 - h}}{{\left( {\frac{{100}}{{\sqrt 3 }}} \right)}}$$      (∵ BD = CE and substituted the value of CE from eq. 1)
$$\eqalign{ & \Rightarrow \left( {100 - h} \right) = \frac{1}{{\sqrt 3 }} \times \frac{{100}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 33.33 \cr & \Rightarrow h = 100 - 33.33 = 66.67\,{\text{m}} \cr} $$
i.e., the height of the pole = 66.67 m

$$\eqalign{ & {\text{in}}\,\Delta PNQ,\,\tan {45^ \circ } = \frac{{PQ}}{{NQ}} \cr & \therefore PQ = NQ = 50{\text{m}} \cr & {\text{in}}\,\Delta MNQ,\,\tan {30^ \circ } = \frac{{MQ}}{{NQ}} \cr & \therefore \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{50}} \cr & \therefore MQ = \frac{{50}}{{\sqrt 3 }} \cr & \therefore h = PM \cr & \,\,\,\,\,\,\,\,\,\, = PQ - MQ \cr & \,\,\,\,\,\,\,\,\,\, = 50 - \frac{{50}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{50}}{{\sqrt 3 }}\left( {\sqrt 3 - 1} \right) \cr} $$
∴ Poster height = $$\frac{{50}}{{\sqrt 3 }}\left( {\sqrt 3 - 1} \right)$$

$$\eqalign{ & {\text{Height}}\,{\text{of}}\,{\text{pole}} = PQ \cr & \tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{m} \cr & \tan {45^ \circ } = 1 = \frac{{PQ}}{n} \cr & {\text{Multiply}}\,{\text{both}}\,{\text{equations}} \cr & \sqrt 3 \times 1 = \frac{{PQ}}{m} \times \frac{{PQ}}{n} \cr & \therefore PQ = \sqrt {mn\sqrt 3 } \,{\text{units}} \cr} $$

$$\eqalign{ & {\text{in}}\,\Delta MNQ,\tan {30^ \circ } = \frac{{MQ}}{{NQ}} \cr & \therefore \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{10}} \cr & \therefore MQ = \frac{{10}}{{\sqrt 3 }} \cr & {\text{Also}}\,{\text{by}}\,{\text{Pythagoras}}\,{\text{theorem}} \cr & M{N^2} = M{Q^2} + N{Q^2} \cr & \therefore {L^2} = \frac{{100}}{3} + 100 \cr & \therefore L = \frac{{20}}{{\sqrt 3 }} \cr & \therefore {\text{Height}}\,{\text{of}}\,{\text{tree}} \cr & = L + MQ \cr & = \frac{{20}}{{\sqrt 3 }} + \frac{{10}}{{\sqrt 3 }} \cr & = \frac{{30}}{{\sqrt 3 }} \cr & = \frac{{3 \times 10}}{{\sqrt 3 }} \cr & = 10\sqrt 3 \,{\text{m}} \cr} $$

$$\eqalign{ & {\text{Original tree height}} = {\text{h}} = {\text{MQ}} \cr & {\text{New}}\,{\text{tree}}\,{\text{height}} = PQ \cr & {\text{in}}\,\Delta MQN,\,\tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{NQ}} \cr & \therefore MQ = \frac{{300}}{{\sqrt 3 }} \cr & {\text{in}}\,\Delta PQN,\,\tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{NQ}} \cr & \therefore PQ = 300\sqrt 3 \cr & {\text{Tree}}\,{\text{grew}} = PQ - MQ \cr & \therefore {\text{Tree}}\,{\text{grew}} \cr & = 300\sqrt 3 - \frac{{300}}{{\sqrt 3 }} \cr & = 300\frac{2}{{\sqrt 3 }} \cr & = 3 \times 100 \times \frac{2}{{\sqrt 3 }} \cr & = 200\sqrt 3 \,{\text{m}} \cr} $$

Let PQ be three and M and N be positions where Mphan stands.
$$\eqalign{ & {\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{MQ}} \cr & \therefore PQ = MQ \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{NQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{10 + MQ}} \cr & \therefore 10 + MQ = \sqrt 3 \,PQ \cr} $$
$$\therefore 10 + PQ = \sqrt 3 \,PQ$$     (As, PQ = MQ)
$$\therefore PQ = \frac{{10}}{{\sqrt 3 - 1}}\,{\text{cm}}$$

$$\eqalign{ & {\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{NQ}} \cr & \therefore {\text{Tree}}\,{\text{height}} \cr & = PQ = NQ \cr & = \left( {240 + MQ} \right) \cr & \tan {60^ \circ } = \sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{MQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240 + MQ}}{{MQ}} \cr & \therefore 240 + MQ = \sqrt 3 \,MQ \cr & \therefore MQ = \frac{{240}}{{\sqrt 3 - 1}}\,{\text{ft}} \cr & \therefore NQ = 240 + MQ \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240\sqrt 3 }}{{\sqrt 3 - 1}}\,{\text{ft}} \cr} $$
= Rohit was this much far away initially

Let the tree break at height h cm from ground at point M.
The broken part makes angle of 30°
∴ Broken Part MP = MN = 24 - h
$$\eqalign{ & {\text{in}}\,\Delta MCQ, \cr & \sin {30^ \circ } = \frac{1}{2} = \frac{{MQ}}{{MN}} = \frac{{\text{h}}}{{24 - {\text{h}}}} \cr & \therefore 24 - {\text{h}} = 2{\text{h}} \cr} $$
∴ h = 8 cm = Tree breaks at this height

In ∠ABR, ∠ARB = 90° and ∠BAR = 45°
Sum of angles of a triangle = 180°
So ∠ABR = 180 - 90 - 45 = 45°
∴ BR = AR
AS = RQ = 10m
Also, BR = BQ - RQ = 25 - 10 = 15m
∴ AR = 15m = Distance between houses