Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

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

Answer: Option C

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

Answer: Option A

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

Answer: Option A

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

Answer: Option A

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

Answer: Option B

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

Answer: Option D

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)
$$\therefore PQ = \frac{{10}}{{\sqrt 3 - 1}}\,{\text{cm}}$$

Answer: Option A

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

Answer: Option B

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

Answer: Option D

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