Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Shadow length = $$\frac{1}{{\sqrt 3 }}$$ Height of man
$$\eqalign{ & \frac{{{\text{Height}}\,{\text{of}}\,{\text{man}}}}{{{\text{Shadow}}\,{\text{length}}}} = \sqrt 3 \cr & \tan \theta = \frac{{{\text{Height}}\,{\text{of}}\,{\text{man}}}}{{{\text{Shadow}}\,{\text{length}}}} = \sqrt 3 \cr & {\text{But}}\,\tan {60^ \circ } = \sqrt 3 \cr} $$
∴ $$\theta $$ = 60° = Angle of elevation of sum

In $$\Delta $$STR, by Pythagoras theorem
$$\eqalign{ & R{T^2} = S{T^2} + R{S^2} \cr & \therefore S{T^2} = {15^2} - {9^2} = 144 \cr & \therefore ST = 12\,{\text{m}} \cr} $$
In $$\Delta $$TQP, by Pythagoras theorem
$$\eqalign{ & P{T^2} = T{Q^2} + P{Q^2} \cr & \therefore T{Q^2} = {15^2} - {12^2} = 81 \cr & \therefore TQ = 9\,{\text{m}} \cr} $$
Distance between houses
⇒ SQ = ST + TQ
⇒ SQ = 12 + 9
⇒ SQ = 21 m

Let MN = Ramesh's fort & PQ = Suresh's fort
From the diagram we can see that
MR = 24 cm & PR = 15 - 8 = 7 cm
By Pythagoras theorem,
Hypotenuse² = (side1)² + (side2)²
$$\eqalign{ & \therefore {\text{MP}} = \sqrt {{{24}^2} + {7^2}} \cr & \therefore {\text{MP}} = 25\,{\text{cm}} \cr} $$

$$\eqalign{ & {\text{In}}\,\Delta PBQ,\,\tan {60^ \circ } = \frac{{PQ}}{{BQ}} \cr & \therefore BQ = \frac{{PQ}}{{\sqrt 3 }} \cr & {\text{In}}\,\Delta PAQ,\,\tan {30^ \circ } = \frac{{PQ}}{{AQ}} \cr & \therefore \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{50 + BQ}} \cr & \therefore PQ = \frac{{50 + BQ}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 + \frac{{PQ}}{{\sqrt 3 }}}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 + PQ}}{{\sqrt 3 \times \sqrt 3 }} \cr & \therefore 3PQ = 50\sqrt 3 + PQ \cr} $$
$$\therefore PQ = 25\sqrt 3 \,{\text{m}} = $$     Height of tree

Angle of Depression = Angle of Elevation
Tower PS = 10 m in height
$$\eqalign{ & {\text{tan}}{45^ \circ } = 1 = \frac{{PS}}{{RS}} \cr & \therefore PS = RS = 10 \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{PS}}{{SQ}} = \frac{{10}}{{SQ}} \cr & \therefore SQ = 10\sqrt 3 \cr & RQ = RS + SQ \cr & \,\,\,\,\,\,\,\,\,\,\, = 10 + 10\sqrt 3 \cr & \,\,\,\,\,\,\,\,\,\,\, = 27.32\,{\text{m}} \cr} $$

$$\eqalign{ & \tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{QR}} \cr & \therefore PQ = \sqrt 3 \,QR \cr & \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{SQ}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{80 + QR}} \cr & \therefore 80 + QR = \sqrt 3 \,PQ \cr & \therefore 80 + QR = 3QR \cr & \therefore QR = 40\,{\text{ft}}{\text{.}} \cr} $$
If we read carefully, we see that Raj (Point R) and the scarecrow (Point Q) are in diagonally opposite corners.
So QR is a diagonal of the square farm.
Diagonal of square = side x$$\sqrt 2 $$
$$\eqalign{ & \therefore 40 = {\text{side}}\,{\text{x}}\sqrt 2 \cr & \therefore {\text{side}} = \frac{{40}}{{\sqrt 2 }} \cr & \therefore {\text{Area}} = {\left( {{\text{side}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{{40}}{{\sqrt 2 }}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{1600}}{2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 800\,{\text{sq}}{\text{.}}\,{\text{ft}}{\text{.}} \cr} $$

Both the man and pole are near each other and are illuminated by same sun from same direction.

So angle of elevation for sun is same for both.
So ration of object to shadow will be same for all objects. (Proportionality Rule)
$$\frac{{{\text{Object}}\,{\text{height}}}}{{{\text{Shadow}}\,{\text{length}}}} = \frac{6}{4} = \frac{{\text{H}}}{{36}}$$
∴ H = 54 ft = Height of pole

$$\eqalign{ & AB = {\text{hill}} = 200{\text{ metre}} \cr & CD = {\text{tower}} \cr & {\text{In }}\Delta \,APC \cr & \tan {30^ \circ } = \frac{{AP}}{{PC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{AP}}{{PC}} \cr & \Rightarrow AP:PC = 1:\sqrt 3 \,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{In }}\Delta \,ABD \cr & \tan {60^ \circ } = \frac{{AB}}{{BD}} \cr & \sqrt 3 = \frac{{AB}}{{BD}} \cr & \Rightarrow AB:BD = \sqrt 3 :1\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & PB = CD{\text{ and }}PC = BD \cr & {\text{Now}} \cr} $$
\[\begin{array}{*{20}{c}} {AB}&:&{BD}&:&{AP} \\ {\sqrt 3 }&:&1&{}&{} \\ {}&{}&{\sqrt 3 }&:&1 \\ 3&:&{\sqrt 3 }&:&1 \end{array}\]
$$\eqalign{ & CD = PB \Rightarrow AB - AP \cr & CD = 3 - 1 \cr & CD = 2{\text{ units}} \cr & AB = 3{\text{ units}} = 200{\text{ metre}} \cr & CD = 2{\text{ units}} = \frac{{200}}{3} \times 2 = 133\frac{1}{3}\,{\text{metre}} \cr} $$

$$\eqalign{ & \Delta ABE \cr & \tan {30^ \circ } = \frac{{AB}}{{EB}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr & \Rightarrow x = \sqrt 3 \times h \cr & \Delta EBD \cr & \tan {60^ \circ } = \frac{{BD}}{{EB}} \cr & \sqrt 3 = \frac{{h + 30 + 30}}{x} \cr & \sqrt 3 = \frac{{h + 60}}{{\sqrt 3 h}} \cr & 3h = h + 60 \cr & 2h = 60 \cr & h = 30{\text{m}} \cr & {\text{Height from water surface}} \cr & = 30 + 30 \cr & = 60{\text{m}} \cr} $$

$$\eqalign{ & {\text{In }}\Delta ABM, \cr & \tan \left( {90 - \theta } \right) = \frac{P}{B} = \frac{{AB}}{{72}} \cr & \frac{1}{{\cot \theta }} = \frac{{72}}{{AB}} \cr & \tan \theta = \frac{{72}}{{AB}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & {\text{In }}\Delta ABN, \cr & \tan \theta = \frac{P}{B} = \frac{{AB}}{{128}}{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{72}}{{AB}} = \frac{{AB}}{{128}} \cr & A{B^2} = 128 \times 72 \cr & AB = \sqrt {16 \times 4 \times 2 \times 36 \times 2} \cr & AB = 4 \times 4 \times 6 \cr & AB = 96 \cr} $$