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Aptitude - Height & Distance Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Height & Distance. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - A man observes the elevation of a balloon to be 45° at a point A .He then walks towards the balloon and at a certain place B finds the elevation to be 60°. He further walks in the direction of the balloon and finds it to be directly over him at a height of 450 m. Distance travelled from A to B is

Answer : A

Explanation

Height & Distance Solution 2

450/BD= tan (60) =>BD =450/√3
450/AD= tan (30) =>AD= 450√3
AD =BD +AB
=>AB=AD-BD= 450√3-450/√3=(450x3-450)/√3=300√3m

Q 2 - A vertical pole fixed to the ground is divided in the ratio 1:4 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 16 m away from the base of the pole, what is the height of the pole?

Answer : C

Explanation

Height & Distance Solution 4

Let CB be the pole and point D divides it such that BD : DC = 1 : 4 = X:4X
Given that AB = 16 m
Let the the two parts subtend equal angles at point A such that
CAD =  BAD = Θ
=>tan  Θ=X/16 =>X=16 tan ( Θ) ------ (1)
=>tan( Θ+  Θ)=4X/16
=>16 tan (2 Θ)=4X
=>16(2tan ( Θ))/(1-tan ( Θ)²)=4X ------ (2)
From eqn 1 & 2 2X/(1-tan ( Θ)²)=4X (X=16tan Θ)
1/(1-(X/16)²)=2
1-(X/16)²=1/2=>16²-
X²=162/2=>X²=128
=>X=8√2
=>Height of pole BC = X+4X=5X=40√2

Q 3 - From the top of mast head of height 210 meters of a ship, a boat is observed at an angle of depression of 30° then the distance between them is

Answer : A

Explanation

Height & Distance Solution 7

From the right angled triangle CAB
Tan(30) =210/X
=>X=210/Tan(30)=210/(1/√3)=210√3

Q 4 - The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

Answer : B

Explanation

Height & Distance Solution 12

Let AB be the building and AC be its shadow.
Then, AC=20m and ∠ACB=60°.Let AB= x m.
Presently AB/AC=tan 60°=√3=>x/10=√3
=>x=10√3m= (10*1.732) m=17.32m.
∴ Height of the building is 17.32m.

Q 5 - From The highest point of a 10 m high building, the edge of rise of the of the highest point of a tower is 60° and the despondency's edge of its foot is 45°,Find The tower's stature. (take√3=1.732)

Answer : D

Explanation

Height & Distance Solution 14

Let AB be the building and CD be the tower.
Draw BE perpendicular to CD.
 At that point CE =AB = 10m, ∠EBD= 60° and ∠ACB= ∠ CBE=45°
AC/AB= cot45°=1 = >AC/10 =1 => AC = 10m.
From △ EBD, we have
DE/BE= tan 60°=√3 => DE/AC= √3
=> DE/10= 1.732 =>DE = 17.3
Height of the tower = CD= CE+DE= (10+17.32) = 27.3 m.

Q 6 - A 10 m long stepping stool is put against a divider. It is slanted at a point of 30°to the ground. The separation of the stepping stool's foot from the divider is:

Answer : C

Explanation

Height & Distance Solution 18

Let AB be the step slanted at 30°to the Ground AC.
Then, AB=10m and

Q 7 - The point of the height of a stepping stool inclining toward a divider is 60°and the step's foot is 7.5 m far from the divider. The stepping stool's length is

Answer : A

Explanation

Height & Distance Solution 19

Let AB be the step inclining toward the divider CB.
Let AC be the flat such that AC=7.5M
What's more, ∠CAB=60°
∴ AB/AC=sec60°=2 => AB/7.5m=2 => AB=15m.
∴ length of the stepping stool is 15m.

Q 8 - From a point on a scaffold over the waterway, the edge of dejection of the banks on inverse sides of the waterway is 30°and 45°respectively. In the event that the scaffold is at tallness of 2.5m from the banks, find the width of the Stream. (Take √3=1.732)

Answer : B

Explanation

Height & Distance Solution 24

Let and B be two point on the banks on inverse sides of the stream.
Let P be a point on the scaffold at stature of 2.5m.
Let PQ-AB.
PQ=2.5m.∠BAP=30°and ∠ABP=45°.
QB/PQ=cot45°=1 => QB/2.5=1 => QB=2.5m.
AQ/PQ =cot30°=√3 => AQ/2.5= √3 => AQ= (2.5)√3m.
Width of the stream =AB= (AQ+QB)=2.5(√3+1)
5/2(1.732+1) m=6.83m.

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