Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & SR = PQ = 2\,m \cr & PS = QR = 10\sqrt 3 \,m \cr & \tan {30^ \circ } = \frac{{TS}}{{PS}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{TS}}{{10\sqrt 3 }} \cr & TS = \frac{{10\sqrt 3 }}{{\sqrt 3 }} = 10\,m \cr & TR = TS + SR = 10 + 2 = 12\,m \cr} $$

Let ED be the taller tower and AB be the shorter tower.
Let C be the point of observation
Given that ∠ ACB = 22.5° and ∠ DCE = 67.5°
Given that C is the midpoint of BD
Hence, BC = CD
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABC, \cr & \tan{22.5^ \circ } = \frac{{AB}}{{BC}}\,.....\left( {eq:1} \right)\, \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,CDE, \cr & \tan {67.5^ \circ } = \frac{{ED}}{{CD}}\,.....\left( {eq:2} \right) \cr & \frac{{\left( {eq:2} \right)}}{{\left( {eq:1} \right)}} \Rightarrow \frac{{\tan {{67.5}^ \circ }}}{{\tan{{22.5}^ \circ }}} = \frac{{\left( {\frac{{ED}}{{CD}}} \right)}}{{\left( {\frac{{AB}}{{BC}}\,} \right)}} \cr & = \frac{{ED}}{{AB}}\,\,\,\,\,\left( {\because CD = BC} \right) \cr & \Rightarrow \frac{{\tan \left( {{{90}^ \circ } - {{22.5}^ \circ }} \right)}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}} \cr & \Rightarrow \frac{{\cot {{22.5}^ \circ }}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}} \cr & \,\,\,\,\,\,\,\,\,\,\left[ {\because \tan \left( {90 - \theta } \right) = \cot \theta } \right] \cr & \Rightarrow \frac{{\left( {\frac{1}{{\tan {{22.5}^ \circ }}}} \right)}}{{\tan {{22.5}^ \circ }}} = \frac{{ED}}{{AB}} \cr & \,\,\,\,\,\,\,\,\,\,\,\left[ {\because \cot \theta = \frac{1}{{\tan \theta }}} \right] \cr & \Rightarrow \frac{{ED}}{{AB}} = \frac{1}{{{{\left( {\tan {{22.5}^ \circ }} \right)}^2}}} \cr & = \frac{1}{{{{\left( {\sqrt 2 - 1} \right)}^2}}} \cr & = {\left( {\frac{1}{{\sqrt 2 - 1}}} \right)^2} \cr & = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}} \right]^2} \cr & = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {2 - 1} \right)}}} \right]^2} \cr & = {\left[ {\frac{{\left( {\sqrt 2 + 1} \right)}}{1}} \right]^2} \cr & = {\left( {\sqrt 2 + 1} \right)^2} \cr & = \left( {2 + 2\sqrt 2 + 1} \right) \cr & = \left( {3 + 2\sqrt 2 } \right) \cr & {\text{Required}}\,{\text{Ratio}} \cr & = ED:AB \cr & = \left( {3 + 2\sqrt 2 } \right):1 \cr} $$

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.
Given that AD = 18 m, ∠ ABC = 60°, ∠ DBC = 30°
Let DC be h
$$\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{{BC}} \cr & h = \frac{{BC}}{{\sqrt 3 }}\,......\left( eq : 1 \right) \cr & \tan {60^ \circ } = \frac{{AC}}{{BC}} \cr & \sqrt 3 = \frac{{18 + h}}{{BC}} \cr & 18 + h = BC \times \sqrt 3 \,......\left( eq: 2 \right) \cr & \frac{eq : 1}{eq : 2} \Rightarrow \frac{h}{{18 + h}} = \frac{{\left( {\frac{{BC}}{{\sqrt 3 }}} \right)}}{{\left( {BC \times \sqrt 3 } \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3} \cr & \Rightarrow 3h = 18 + h \cr & \Rightarrow 2h = 18 \cr & \Rightarrow h = 9\,{\text{m}} \cr} $$
i.e., the height of the tower = 9 m

Let AB be the man and CD be the window
Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
∠ DAF = 45°, ∠ CAF = 60°
From the diagram, AF = BE = 5 m
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFD, \cr & \tan {45^ \circ } = \frac{{DF}}{{AF}} \cr & 1 = \frac{{DF}}{5} \cr & DF = 5\,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,AFC, \cr & \tan {60^ \circ } = \frac{{CF}}{{AF}} \cr & \sqrt 3 = \frac{{CF}}{5} \cr & CF = 5\sqrt 3 \,......\,\left( 2 \right) \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{window}} \cr & = CD = \left( {CF - DF} \right) \cr} $$
  $$ = 5\sqrt 3 - 5$$     [∵ Substituted the value of CF and DF from (1) and (2)]
$$\eqalign{ & = 5\left( {\sqrt 3 - 1} \right) \cr & = 5\left( {1.73 - 1} \right) \cr & = 5 \times 0.73 \cr & = 3.65\,{\text{m}} \cr} $$

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m, ∠ ABD = 30°, ∠ ACD = 60°,
Let CD = x, AD = h
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,CDA \cr & \tan {60^ \circ } = \frac{{AD}}{{CD}} \cr & \sqrt 3 = \frac{h}{x}\,......\left( {eq:1} \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,BDA \cr & \tan {30^ \circ } = \frac{{AD}}{{BD}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{{70 + x}}\,......\left( {eq:2} \right) \cr & \frac{{eq:1}}{{eq:2}} \Rightarrow \frac{{\sqrt 3 }}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}} = \frac{{\left( {\frac{h}{x}} \right)}}{{\left( {\frac{h}{{70 + x}}} \right)}} \cr & \Rightarrow 3 = \frac{{70 + x}}{x} \cr & \Rightarrow 2x = 70 \cr & \Rightarrow x = 35 \cr} $$
Substituting this value of x in eq : 1, we have
$$\eqalign{ & \sqrt 3 = \frac{h}{{35}} \cr & \Rightarrow h = 35\sqrt 3 = 35 \times 1.73 \cr & = 60.55 \approx 60.6 \cr} $$

$$\eqalign{ & \tan {30^ \circ } = \frac{{RQ}}{{PQ}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{200}}{{PQ}} \cr & PQ = 200\sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\, = 200 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\, = 346\,{\text{m}} \cr} $$

Let DC be the wall, AB be the tree.
Given that ∠DBC = 30°, ∠DAE = 60°, DC = 11 m
$$\eqalign{ & \tan {30^ \circ } = \frac{{DC}}{{BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{11}}{{BC}} \cr & BC = 11\sqrt 3 \,m \cr & AE = BC = 11\sqrt 3 \,m\,.....\left( 1 \right) \cr & \tan {60^ \circ } = \frac{{ED}}{{AE}} \cr} $$
$$\sqrt 3 = \frac{{ED}}{{11\sqrt 3 }}$$   [∵ Substituted value of AE from (1)]
$$\eqalign{ & ED = 11\sqrt 3 \times \sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\, = 11 \times 3 \cr & \,\,\,\,\,\,\,\,\,\, = 33 \cr & {\text{Height}}\,{\text{of}}\,{\text{the}}\,{\text{tree}} \cr & = AB = EC = \left( {ED + DC} \right) \cr & = 33 + 11 \cr & = 44\,{\text{m}} \cr} $$

Consider a right-angled triangle PQR as shown below.

Let QR = 3 and PR = 5 such that
$$\sin \theta = \frac{3}{5}\,\,\,\,\left[ {{\text{i}}{\text{.e}}{\text{.}},\theta = {{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right]$$
$$PQ = \sqrt {P{R^2} - Q{R^2}} $$     (∵ Pythagorean theorem)
$$\eqalign{ & = \sqrt {{5^2} - {3^2}} \cr & = 4 \cr} $$
$${\text{i}}{\text{.e}}{\text{.}},\,{\text{when}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right),$$     PQ : QR = 4 : 3 ......(eq : 1)

Now Let's solve the question. Let P be the point of observation and QR be the tower as shown in the below diagram.

$${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$     and PQ = 20 m
We know that PQ : QR = 4 : 3 (from eq : 1)
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}},\,20:QR = 4:3 \cr & \Rightarrow 20 \times 3 = QR \times 4 \cr & \Rightarrow QR = 15\,{\text{m}} \cr} $$
i.e. height of the tower = 15 m

---

Solution 2

Let P be the point of observation and QR be the tower.
$${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$     and PQ = 20 m
Let the height of the tower, QR = h and PR = x
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta PQR, \cr & \sin \theta = \frac{{QR}}{{PR}} \cr & \Rightarrow \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right] = \frac{h}{x} \cr & \Rightarrow \frac{3}{5} = \frac{h}{x} \cr & \Rightarrow x = \frac{{5h}}{3}\,.....\left( {eq:1} \right) \cr} $$
From Pythagorean theorem, we have
$$\eqalign{ & P{Q^2} + Q{R^2} = P{R^2} \cr & {20^2} + {h^2} = {x^2} \cr} $$
$${20^2} + {h^2} = {\left( {\frac{{5h}}{3}} \right)^2}$$   (∵ Substituted the value of x from eq:1)
$$\eqalign{ & {20^2} + {h^2} = \frac{{25{h^2}}}{9} \cr & \frac{{16{h^2}}}{9} = {20^2} \cr & \frac{{4h}}{3} = 20 \cr & h = \frac{{3 \times 20}}{4} = 15{\text{m}} \cr} $$
i.e. height of the tower = 15 m

Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC
Given that BA = 10 m, ∠ BAC = 60°
$$\eqalign{ & \sin {60^ \circ } = \frac{{BC}}{{BA}} \cr & \frac{{\sqrt 3 }}{2} = \frac{{BC}}{{10}} \cr & BC = 10 \times \frac{{\sqrt 3 }}{2} \cr & \,\,\,\,\,\,\,\,\,\, = 5 \times 1.73 \cr & \,\,\,\,\,\,\,\,\,\, = 8.65\,{\text{m}} \cr} $$

$$\eqalign{ & \tan {45^ \circ } = \frac{{SR}}{{QR}} \cr & \tan {30^ \circ } = \frac{{SR}}{{PR}} = \frac{{SR}}{{\left( {PQ + QR} \right)}} \cr} $$
Two equations and 3 variables. Hence we can not find the required value with the given data.

(Note that if one of SR, PQ, QR is known, this becomes two equations and two variables and if that was the case, we could have found out the required value.)