Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option C

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, ∠ACB = 30° and ∠ADB = 45°
$$\frac{{AB}}{{AC}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ \Rightarrow AC = AB \times \sqrt 3 =100 \sqrt 3{\text{m}}$$
$$\frac{{AB}}{{AD}} = \tan {45^ \circ } = 1$$
$$ \Rightarrow AD = AB = 100{\text{m}}$$
∴ CD = (AC+AD) = (100√3 +100)
= 100(√3 +1) = 100(1.73+1) =100 × 2.73 = 273m

Answer: Option D

One of AB, AD and CD must have given.

So, the data is inadequate.

Answer: Option D

Let AB be the wall and BC be the ladder.

Then, ∠ACB = 60° = AC = 4.6m
$$\frac{{AC}}{{BC}} = \cos {60^ \circ } = \frac{1}{2}$$
⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

Answer: Option A

Let AB be the observer and CD be the tower.


Draw BE ⊥ CD
Then CE = AB = 1.6m
BE = AC = 20√3m
$$\frac{{DE}}{{BE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ \Rightarrow DE = \frac{{BE}}{{\sqrt 3 }} = \frac{{20\sqrt 3 }}{{\sqrt 3 }} = 20$$
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m

Answer: Option C

Let AB be the tower.

Then, ∠APB = 30° and AB = 100m
$$\frac{{AB}}{{AP}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
⇒ AP = AB × √3 = 100 × √3
⇒ AP = 100 × 1.73 = 173m

Answer: Option A

Let AB be the tree and AC be its shadow.

Let ∠ACB = θ
$$\frac{{AC}}{{AB}} = \cot \theta $$
$$\frac{{AC}}{{AB}} = \sqrt 3 $$
$$\cot {30^ \circ } = \sqrt 3 $$
∴ θ = 30°

Answer: Option C

Let AB be the tower and C and D be the points of observation.
Then, ∠ACB = 30°, ∠ADB = 45° and CD = 20m
Let AB = h then,
$$\eqalign{ & \frac{{AB}}{{AC}} = \tan 30^\circ = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow AC = AB \times \sqrt 3 = h\sqrt 3 {\kern 1pt} {\text{And,}} \cr & \Rightarrow \frac{{AB}}{{AD}} = \tan 45^\circ = 1 \cr & \Rightarrow AD = AB = h \cr & \, \, \, \, \, CD = 20 \cr & \Rightarrow \left( {AC - AD} \right) = 20 \cr & \Rightarrow h\sqrt 3 - h = 20 \cr & \therefore h = \frac{{20}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & = 10\left( {\sqrt 3 + 1} \right){\text{m}} \cr & = \left( {10 \times 2.73} \right){\text{m}} \cr & = 27.3 {\text{m}} \cr} $$

Answer: Option A

Let AB be the tower and C and D be the objects.
Then, AB = 150 m, ∠ACB = 45° and ∠ADB = 60°
$$\eqalign{ & \frac{{AB}}{{AD}} = \tan {60^ \circ } = \sqrt 3 \cr & \Rightarrow AD = \frac{{AB}}{{\sqrt 3 }} = \frac{{150}}{{\sqrt 3 }} \cr & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow AC = AB = 150{\text{ m}} \cr & \therefore CD = \left( {AC - AD} \right) \cr & = \left( {150 - \frac{{150}}{{\sqrt 3 }}} \right){\text{m}} \cr & = \left[ {\frac{{150\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }}} \right]{\text{m}} \cr & = 50\left( {3 - \sqrt 3 } \right){\text{m}} \cr & = \left( {50 \times 1.27} \right){\text{m}} \cr & = 63.5\,{\text{m}} \cr} $$

Answer: Option B

Length of the tower AB = h meter.
$$\eqalign{ & \angle DAC = \angle ACB = {60^ \circ } \cr & BC = 70{\text{ meter}} \cr & {\text{In }}\vartriangle {\text{ABC,}} \cr & {\text{tan }}{60^ \circ } = \frac{{AB}}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{h}{{70}} \cr & \Rightarrow h = 70\sqrt 3 {\text{ meter}} \cr} $$

Answer: Option C

Let AB be the tower and C and D be the two positions of the car.
Then, \[\angle ACB = {45^ \circ },\]    \[\angle ADB = {30^ \circ }\]
Let, AB = h, CD = x and AC = y
$$\eqalign{ & \frac{{AB}}{{AC}} = \tan {45^ \circ } = 1 \cr & \Rightarrow \frac{h}{y} = 1 \cr & \Rightarrow y = h \cr & \frac{{AB}}{{AD}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{h}{{x + y}} = \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x + y = \sqrt 3 h \cr & \therefore x = \left( {x + y - y} \right) \cr & \,\,\,\,\,\,\, = \sqrt 3 h - h \cr & \,\,\,\,\,\,\, = h\left( {\sqrt 3 - 1} \right) \cr} $$
Now, $$h\left( {\sqrt 3 - 1} \right)$$   is covered in 12 min.
So, h will be covered in→
$$\eqalign{ & \left[ {\frac{{12}}{{h\left( {\sqrt 3 - 1} \right)}} \times h} \right] \cr & = \frac{{12}}{{\left( {\sqrt 3 - 1} \right)}}\min \cr & = \left( {\frac{{1200}}{{73}}} \right)\min \cr & = 16\min ,\,\,23\sec \cr} $$