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Height And Distance - Aptitude MCQ Questions and Solutions with Explanations

Height And Distance MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Height And Distance with easy and logical explanations. Height And Distance MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-10 section-1

Each section contains 100 MCQs on Height and Distance .
91
As observed from the top a light house, 120√3 m above the sea level, the angle of depression of a ship sailing towards it changes from 30° to 60°. The distance travelled by the during the period of observation is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Height and Distance mcq question image
Height and Distance mcq question image
$$\eqalign{ & \sqrt 3 \to 120\sqrt 3 \cr & 1{\text{u}} \to 120 \cr & 2{\text{u}} \to 120 \times 2 = 240{\text{m}} \cr} $$
92
From the top of hill 240 m high, the angles the angles of depression of the top and bottom of a pole are 30° and 60°, respectively. The difference (in m) between the height of the pole and its distance from the hill is:
Discuss
Answer & Solution
Answer: Option D
Solution:
Height and Distance mcq question image
$$\eqalign{ & 3 \to 240 \cr & \sqrt 3 \to \frac{{240}}{3} \times \sqrt 3 = 80\sqrt 3 \cr & {\text{Height of the pole}} \cr & = 160 - 80\sqrt 3 \cr & = 80\left( {2 - \sqrt 3 } \right) \cr} $$
93
A person was standing on a road near a mall. He was 1215 m away from the mall and able to see the top of the mall from the road in such a ways that the top of a tree, which is in between him and the mall, was exactly in line of sight with the top of the mall. The tree height is 20 m and it is 60 m away from him. How tall (in m) is the mall?
Discuss
Answer & Solution
Answer: Option C
Solution:
Height and Distance mcq question image
$$\eqalign{ & \tan \theta = \frac{{20}}{{60}} = \frac{1}{3} \cr & {\text{Now }}\tan \theta = \frac{h}{{1215}} \cr & \frac{1}{3} = \frac{h}{{1215}} \cr & h = 405 \cr} $$
94
The distance between two parallel poles is 40√3 m. The angle of depression of the top of the second pole when seen from the top of first pole is 30°. What will be the height of second tower if the first pole is 100 m long
Discuss
Answer & Solution
Answer: Option D
Solution:
Height and Distance mcq question image
$$\eqalign{ & \tan {30^ \circ } = \frac{x}{{40\sqrt 3 }} \cr & \frac{1}{{\sqrt 3 }} = \frac{x}{{40\sqrt 3 }} \cr & x = 40 \cr & {\text{So, }}y = 100 - 40 = 60 \cr} $$
95
A ladder is resting against a wall, the angle between the foot of the ladder and the wall is 45° and the foot of the ladder is 6.6 m away from the wall. The length of the ladder (in m) is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Height and Distance mcq question image
Length of the ladder = 6.6√2 m
96
P and Q are two points on the ground on either side of a pole. The angles of elevation of the top of the pole as observed from P and Q are 60° and 30°, respectively and the distance between them is 84√3 m. What is the height (in m) of the pole?
Discuss
Answer & Solution
Answer: Option B
Solution:
Height and Distance mcq question image
\[{60^ \circ } = \frac{{\sqrt 3 }}{1}\begin{array}{*{20}{c}} \to &h \\ \to &{\frac{h}{{\sqrt 3 }}} \end{array}\]
$$PN = \frac{h}{{\sqrt 3 }}$$
\[{30^ \circ } = \frac{1}{{\sqrt 3 }}\begin{array}{*{20}{c}} \to &h \\ \to &{\sqrt 3 h} \end{array}\]
$$\eqalign{ & NQ = \sqrt 3 h \cr & PN + NQ = 84\sqrt 3 \cr & \frac{h}{{\sqrt 3 }} + \sqrt 3 h = 84\sqrt 3 \cr & h\left( {1 + 3} \right) = 84 \times \sqrt 3 \times \sqrt 3 \cr & h = \frac{{84 \times 3}}{4} \cr & h = 63\,{\text{m}} \cr} $$
97
A ladder is placed along a wall such that its upper end is touching the top of the wall. The foot of the ladder is 10 ft away from the wall and the ladder is making an angle of 60° with the ground. When a man starts climbing on it, it slips and now ladder makes an angle of 30° with ground. How much did the ladder slip?
Discuss
Answer & Solution
Answer: Option A
Solution:
Height and Distance mcq question image
$$\eqalign{ & AB = 10\tan {60^ \circ } = 10\sqrt 3 \cr & BC = 10\sec {60^ \circ } = 20 \cr} $$
Since, length of ladder will remain same, hence BC = DE = 20
$$\eqalign{ & {\text{In }}\Delta ADE,\,DE = 20{\text{ ft}} \cr & \Rightarrow \frac{{AD}}{{DE}} = \cos {30^ \circ } \cr & \Rightarrow AD = 20 \times \frac{{\sqrt 3 }}{2} = 10\sqrt 3 \cr & \Rightarrow x = AD - AC \cr & \Rightarrow x = 10\sqrt 3 - 10 \cr & \Rightarrow x = 10\left( {\sqrt 3 - 1} \right){\text{ft}} \cr} $$
98
From the top of a 12 m high building, The angle of elevation of the top of a tower is 60° and the angle of depression of the foot of the tower is θ. Such that tan θ = $$\frac{3}{4}$$, what is the height of the tower (√3 = 1.73)?
Discuss
Answer & Solution
Answer: Option A
Solution:
Height and Distance mcq question image
Let AB be the building and EC be the tower.
$$\eqalign{ & {\text{In }}\Delta ADE \cr & \tan {60^ \circ } = \frac{{ED}}{{AD}} \cr & \sqrt 3 = \frac{{ED}}{{16}} \cr & ED = 16\sqrt 3 = 16 \times 1.73 = 27.68\,{\text{m}} \cr} $$
Hence the height of the tower
EC = CD + ED
= 12 + 27.68
= 39.68 m
99
A poster is on top of a building. A person is standing on the ground at a distance of 50 m from the building. The angles of elevation to the top of the poster and bottom of the poster are 45° and 30°, respectively. What is 200% of the height (in cm) of the poster?
Discuss
Answer & Solution
Answer: Option D
Solution:
Height and Distance mcq question image
$$\eqalign{ & \tan {30^ \circ } = \frac{{BD}}{{BC}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{BD}}{{50}} \cr & \Rightarrow BD = \frac{{50}}{{\sqrt 3 }} \cr & \tan {45^ \circ } = \frac{{AB}}{{BC}} \cr & \Rightarrow 1 = \frac{{AB}}{{50}} \cr & \Rightarrow AB = 50 \cr & \therefore AD = AB - BD \cr & = 50 - \frac{{50}}{{\sqrt 3 }} \cr & = \frac{{50\left( {\sqrt 3 - 1} \right)\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} \cr & = \frac{{50}}{3}\left( {3 - \sqrt 3 } \right){\text{m}} \cr & 200\% {\text{ of }}AD = \frac{{50}}{3}\left( {3 - \sqrt 3 } \right) \times \frac{{200}}{{100}} \cr & = \frac{{100}}{3}\left( {3 - \sqrt 3 } \right){\text{m}} \cr} $$
100
From the top and bottom of a straight hill; the angle of depression and elevation of the top of a pillar of 10 m height are observed to be 60° and 30° respectively. The height (in metres) of the hill is
Discuss
Answer & Solution
Answer: Option D
Solution:
According to the question
Height and Distance mcq question image
From figure, AB = CD
Height of pillar BC = AD = 10 m
In ΔABC,
AB = BC cot 30°
AB = 10√3 m
Also AB = CD = 10√3 m
In ΔCDE,
DE = DC tan 60°
DE = 10√3 √3 = 30
Height of the hill = AD + DE
= 10 + 30
= 40 m