Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that ∠ CAD = ∠ BAD = $$\theta $$
From "Angle Bisector Theorem", we have
$$\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}$$
$$ \Rightarrow \frac{1}{9} = \frac{{15}}{{AC}}$$     [∵ BD : DC = 1 : 9 and AB = 15(given)]
$$ \Rightarrow AC = 15 \times 9\,{\text{m}}\,......\left( {eq:1} \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC,$$
$$CB = \sqrt {A{C^2} - A{B^2}} $$     (∵ Pythagorean theorem)
$$ = \sqrt {{{\left( {15 \times 9} \right)}^2} - {{15}^2}} $$     [∵AC=15 × 9(eq : 1) and AB=15 m(given)]
$$\eqalign{ & = \sqrt {{{15}^2} \times {9^2} - {{15}^2}} \cr & = \sqrt {{{15}^2}\left( {{9^2} - 1} \right)} \cr & = \sqrt {{{15}^2} \times 80} \cr & = \sqrt {{{15}^2} \times 16 \times 5} \cr & = 15 \times 4 \times \sqrt 5 \cr & = 60\sqrt 5 \,{\text{m}} \cr} $$

Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m
$$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr & \Rightarrow 1 = \frac{{AB}}{{100}} \cr & \Rightarrow AB = 100\,......\left( {eq:1} \right) \cr} $$
$$\tan {30^ \circ } = \frac{{AB}}{{BD}}$$
$$ \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{100}}{{BD}}$$     (∵ Substituted the value of AB from equation 1)
$$ \Rightarrow BD = 100\sqrt 3 $$
$$\eqalign{ & CD = \left( {BD - BC} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {100\sqrt 3 - 100} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 100\left( {\sqrt 3 - 1} \right) \cr} $$
It is given that the distance CD is covered in 10 seconds.
i.e., the distance $$100\left( {\sqrt 3 - 1} \right)$$     is covered in 10 seconds.
$$\eqalign{ & {\text{Required}}\,{\text{speed}} \cr & = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{10}} \cr & = 10\left( {1.73 - 1} \right) \cr & = 7.3\,{\text{meter/seconds}} \cr & = 7.3 \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = 26.28\,{\text{km/hr}} \cr} $$

Let C and D be the position of the aeroplanes.
Given that CB = 900 m, ∠ CAB = 60°, ∠ DAB = 45°
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC, \cr & \tan {60^ \circ } = \frac{{CB}}{{AB}} \cr & \sqrt 3 = \frac{{900}}{{AB}} \cr & AB = \frac{{900}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{900 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{900\sqrt 3 }}{3} \cr & \,\,\,\,\,\,\,\,\,\, = 300\sqrt 3 \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABD, \cr & \tan {45^ \circ } = \frac{{DB}}{{AB}} \cr & 1 = \frac{{DB}}{{AB}} \cr & DB = AB = 300\sqrt 3 \cr & \cr & {\text{Required}}\,{\text{height}} \cr & = CD \cr & = \left( {CB - DB} \right) \cr & = \left( {900 - 300\sqrt 3 } \right) \cr & = \left( {900 - 300 \times 1.73} \right) \cr & = \left( {900 - 519} \right) \cr & = 381\,{\text{m}} \cr} $$

Consider the diagram shown above where QR represents the tree and PQ represents its shadow
$$\eqalign{ & {\text{We}}\,{\text{have}},\,QR = PQ \cr & {\text{Let}}\,\angle QPR = \theta \cr} $$
$$\tan \theta = \frac{{QR}}{{PQ}} = 1$$     (since QR = PQ)
$$ \Rightarrow \theta = {45^ \circ }$$
i.e., required angle of elevation = 45°

Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, ∠ BAD = 30°, ∠ BCD = 60°
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABD, \cr & \tan {30^ \circ } = \frac{{BD}}{{BA}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{50}}{{BA}} \cr & \Rightarrow BA = 50\sqrt 3 \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CBD, \cr & \tan {60^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{{50}}{{BC}} \cr & \Rightarrow BC = \frac{{50}}{{\sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 }}{3} \cr} $$
Distance between the two persons
$$\eqalign{ & = AC = BA + BC \cr & = 50\sqrt 3 + \frac{{50\sqrt 3 }}{3} \cr & = \sqrt 3 \left( {50 + \frac{{50}}{3}} \right) \cr & = \frac{{200\sqrt 3 }}{3}\,{\text{m}} \cr} $$
i.e., the distance travelled by the car in 10 seconds = $$\frac{{200\sqrt 3 }}{3}\,{\text{m}}$$
$$\eqalign{ & {\text{Speed}}\,{\text{of the car}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & = \frac{{\left( {\frac{{200\sqrt 3 }}{3}} \right)}}{{10}} = \frac{{200\sqrt 3 }}{3}\,{\text{m/s}} \cr & = \frac{{200\sqrt 3 }}{3} \times \frac{{18}}{5}\,{\text{km/hr}} \cr & = {\text{24}}\sqrt 3 \,{\text{km/hr}} \cr} $$

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, ∠ ADC = 30° , ∠ ACB = 45°
Let AB = h, BC = x, CD = y
$$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{x} \cr & \Rightarrow 1 = \frac{h}{x} \cr & \Rightarrow h = x\,......\left( 1 \right) \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AB}}{{\left( {BC + CD} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{x + y}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{x + y}} \cr & \Rightarrow x + y = \sqrt 3 \,h \cr & \Rightarrow y = \sqrt 3 \,h - x \cr} $$
$$ \Rightarrow y = \sqrt 3 \,h - h$$     (∵ Substituted the value of x from equation 1 )
$$ \Rightarrow y = h\left( {\sqrt 3 - 1} \right)$$
Given that distance y is covered in 8 minutes.
i.e, distance $$h\left( {\sqrt 3 - 1} \right)$$   is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
$$\eqalign{ & h\left( {\sqrt 3 - 1} \right) \propto 8\,......\left( {\text{A}} \right) \cr & h \propto t\,..............\left( {\text{B}} \right) \cr & \frac{{\left( {\text{A}} \right)}}{{\left( {\text{B}} \right)}} \Rightarrow \frac{{h\left( {\sqrt 3 - 1} \right)}}{h} = \frac{8}{t} \cr & \Rightarrow \left( {\sqrt 3 - 1} \right) = \frac{8}{t} \cr & \Rightarrow t = \frac{8}{{\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{\left( {1.73 - 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{.73}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{800}}{{73}}\,{\text{minutes}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 10\frac{{70}}{{73}}\,{\text{minutes}} \cr & \approx {\text{10}}\,{\text{minutes}}\,{\text{57}}\,{\text{seconds}} \cr} $$

Let AB and CD be the poles with heights h and 2h respectively.
Given that distance between the poles, BD = 200 m
Let E be the middle point of BD,
∠ AEB = $$\theta $$
∠ CED = (90 - $$\theta $$)   (∵ given that angular elevations are complementary)
Since E is the middle point of BD, we have BE = ED = 100 m
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABE, \cr & \tan \theta = \frac{{AB}}{{BE}} \cr & \tan \theta = \frac{h}{{100}} \cr & h = 100\tan \theta \,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta EDC, \cr & \tan\left( {90 - \theta } \right) = \frac{{CD}}{{ED}} \cr} $$
$$\cot \theta = \frac{{2h}}{{100}}$$     $$\left[ {\because \tan \left( {90 - \theta } \right) = \cot \theta } \right]$$
$$\eqalign{ & 2h = 100\cot \theta \,......\left( 2 \right) \cr & \left( 1 \right) \times \left( 2 \right) \cr} $$
$$ \Rightarrow 2{h^2} = {100^2}$$     $$\left[ {\because \tan \theta \times \cot\theta = \tan\theta \times \frac{1}{{\tan \theta }} = 1} \right]$$
$$\eqalign{ & \Rightarrow \sqrt 2 \,h = 100 \cr & \Rightarrow h = \frac{{100}}{{\sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 50\sqrt 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 50 \times 1.41 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 70.5 \cr & \Rightarrow 2h = 2 \times 70.5 = 141 \cr} $$
i.e. the height of the poles are 70.5 m and 141 m.

Let ED be the building and AC be the tower.
Given that ED = 230 m, ∠ADC = b, ∠AEB = a
Also given that tan a = $$\frac{5}{{12}}$$ and tan b = $$\frac{4}{{5}}$$
Let AC = h
Required Distance = Distance between the top of these buildings = AE
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABE, \cr & \tan \left( a \right) = \frac{{AB}}{{BE}} \cr} $$
$$ \Rightarrow \frac{5}{{12}} = \frac{{\left( {h - 230} \right)}}{{BE}}$$     [∵ tan(a) = $$\frac{5}{{12}}$$ (given), AB = (AC - BC) = (AC - ED) = (h - 230)]
$$\eqalign{ & \Rightarrow BE = \frac{{12\left( {h - 230} \right)}}{5}\,.....\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ACD, \cr & \tan \left( b \right) = \frac{{AC}}{{CD}} \cr} $$
$$ \Rightarrow \frac{4}{5} = \frac{h}{{CD}}$$    [∵ tan(b) = $$\frac{4}{5}$$ (given), AC = h]
$$\eqalign{ & \Rightarrow CD = \frac{{5h}}{4}\,.....\left( 2 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{diagram,}}\,BE = CD \cr} $$
$$ \Rightarrow \frac{{12\left( {h - 230} \right)}}{5} = \frac{{5h}}{4}$$     (from equation 1 & 2)
$$\eqalign{ & \Rightarrow 48h - \left( {4 \times 12 \times 230} \right) = 25h \cr & \Rightarrow 23h = \left( {4 \times 12 \times 230} \right) \cr & \Rightarrow h = \frac{{\left( {4 \times 12 \times 230} \right)}}{{23}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 480\,{\text{m}}\,......\left( 3 \right) \cr & AB = \left( {AC - BC} \right) \cr} $$
$$\,\,\,\,\,\,\,\,\,\,\,\, = \left( {480 - 230} \right)$$    [∵ Since AC = h = 480 (from 3) and BC = ED = 230 m(given)]
$$\,\,\,\,\,\,\,\,\,\,\,\, = 250\,{\text{m}}$$
In the triangle ABE, tan(a) =$$\frac{5}{{12}}$$ . Let's figure out the value of sin(a) now.
Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = $$\frac{5}{{12}}$$
$$\eqalign{ & {\text{hypotenuse}} = \sqrt {{5^2} + {{12}^2}} = 13 \cr & {\text{i}}{\text{.e}}{\text{.,}} \cr & \sin \left( a \right) = \frac{{{\text{opposite}}\,{\text{side}}}}{{{\text{hypotenuse}}}} = \frac{5}{{13}} \cr} $$
We have seen that $$\sin \left( a \right) = \frac{5}{{13}}$$
$$\eqalign{ & \Rightarrow \frac{{AB}}{{AE}} = \frac{5}{{13}} \cr & \Rightarrow AE = AB \times \frac{{13}}{5} \cr & = 250 \times \frac{{13}}{5} \cr & = 650\,{\text{m}} \cr} $$

i.e., Distance between the top of the buildings = 650 m

Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.
Given that height of the tower, AC = 80 m and the angle of depression, ∠DAB = 30°
∠ABC = ∠DAB = 30° (because DA || BC)
$$\eqalign{ & \tan {30^ \circ } = \frac{{AC}}{{BC}} \cr & \Rightarrow \tan {30^ \circ } = \frac{{80}}{{BC}} \cr & \Rightarrow BC = \frac{{80}}{{\tan {{30}^ \circ }}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{80}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}} \cr & = 80 \times 1.73 = 138.4\,{\text{m}} \cr} $$
i.e., Distance of the bus from the foot of the tower = 138.4 m

Let A be the foot and C be the summit of a mountain.
Given that ∠CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x
Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that ∠DAY = 30°
It is also given that from the point D, the elevation is 60°
i.e., ∠CDE = 60°
$$\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC, \cr & \tan {45^ \circ } = \frac{{CB}}{{AB}} \cr} $$
$$ \Rightarrow 1 = \frac{x}{{AB}}$$    [∵ CB = x(the height of the mountain)]
$$\eqalign{ & \Rightarrow AB = x\,......\left( 1 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta AYD, \cr & \sin {30^ \circ } = \frac{{DY}}{{AD}} \cr} $$
$$ \Rightarrow \frac{1}{2} = \frac{{DY}}{2}$$    (∵ Given that AD = 2)
$$\eqalign{ & \Rightarrow DY = 1\,.......\left( 2 \right) \cr & \cos {30^ \circ } = \frac{{AY}}{{AD}} \cr} $$
$$ \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{AY}}{2}$$    (∵ Given that AD = 2)
$$\eqalign{ & \Rightarrow AY = \sqrt 3 \,......\left( 3 \right) \cr & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CED, \cr & \tan {60^ \circ } = \frac{{CE}}{{DE}} \cr} $$
$$ \Rightarrow \tan {60^ \circ } = \frac{{\left( {CB - EB} \right)}}{{YB}}$$     [∵ CE = (CB - EB) and DE = YB)]
$$ \Rightarrow \tan {60^ \circ } = \frac{{\left( {CB - DY} \right)}}{{AB - AY}}$$     [∵ EB = DY and YB = (AB - AY)]
$$ \Rightarrow \tan {60^ \circ } = \frac{{\left( {x - 1} \right)}}{{\left( {x - \sqrt 3 } \right)}}$$
[∵ CB = x, DY = 1(eq:2), AB = x(eq:1) and AY = $${\sqrt 3 }$$ (eq:3)]
$$\eqalign{ & \Rightarrow \sqrt 3 = \frac{{\left( {x - 1} \right)}}{{\left( {x - \sqrt 3 } \right)}} \cr & \Rightarrow x\sqrt 3 - 3 = x - 1 \cr & \Rightarrow x\left( {\sqrt 3 - 1} \right) = 2 \cr & \Rightarrow 0.73x = 2 \cr & \Rightarrow x = \frac{2}{{0.73}} = 2.7 \cr} $$

i.e., the height of the mountain = 2.7 km