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.

Questions and Answers

Q 1 - The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

A - 15√3

B - 7.5

C - 15√2

D - 7.5√2

Answer : D


Height & Distance Solution 1

Let AB be the wall and BC be the ladder.
Then, ∠ACB = 45° and AC = 7.5 m
AC/BC= Cos (45) =1/√2

Q 2 - An aeroplane when 750 m high passes vertically above another aeroplane at an instant when their angles of elevation at same observing point are 45° and 30&de; respectively. Approximately, how many meters higher is the one than the other?

A - 250(1- √3)

B - 750(3- √3)

C - 250(3- √3)

D - 275(1- √3)

Answer : C


Height & Distance Solution 5

Let C and D be the position of the aeroplanes.
Given that CB = 900 m,∠CAB = 60°,∠DAB = 45°
From the right △ ABC,
From the right △ ADB,
=> Required height CD=CB-DB=750-750/√3=250(3- √3)

Q 3 - A man in a boat is rowing away from a cliff (180 meters high), take 90 seconds to change angle of elevation of the top of cliff from 30° to 45°.The speed of the boat is

A - 2(√3) m/sec

B - 3(√3-1) m/sec

C - 2(√3-1) m/sec

D - (√3-1) m/sec

Answer : C


Height & Distance Solution 8

From right angled triangle ADB,
Tan45=AB/AD =>AB=AD=180
 From right angled triangle ACB,
Tan 30=180/(CD+180)
Speed =Distance/Time=180(√3-1)/90=2(√3-1) m/sec

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.

A - 16.32m

B - 17.32 m

C - 18.32 m

D - 19.32m

Answer : B


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 - Two men are inverse sides of a tower. They gauge the edge of the rise of the highest point of the tower as 30° and 45° respectively. On the off chance that the tallness of the tower is 50 m, discover the separation between the two men. (Take √3=1.732)

A - 135.5m

B - 136.5 m

C - 137.5 m

D - 138.5m

Answer : B


Height & Distance Solution 15

Let AB be the tower and let C and D be the two's positions men.
At that point ∠ACB=30°,∠ADB= 45°and AB= 50 m
AC/AB = Cot30°=√3 => AC/50 = √3
AD/AB=cot 45°=1 => AD/50=1
=> AD=50M.
Separation between the two men =CD= (AC+AD)
= (50√3+50) m=50(√3+1)

Q 6 - At a moment, the shadow's length of a shaft is √3 times the stature of the shaft. The edge of rise of the sun is:

A - 30°

B - 45°

C - 60°

D - 75°

Answer : A


Height & Distance Solution 17

Let AB be the post and AC be its shadow.
Let AB =X m. Then, AC= √3Xm.Let∠ACB=θ.
AB/AC=tanθ => tanθ = X/√3X = 1/√3 =tan30°.
∴ θ = 30°.
Hence, the point of rise is 30°.

Q 7 - A kite is flying at a tallness of 75 m from the level of ground, joined to a string slanted at 60° to the level. The string's length is:

A - 50 √2 m

B - 50√3 m

C - 50m/√2

D - 50m/√3

Answer : B


Height & Distance Solution 20

Let AB be the kite and AC be the level ground
So that BC - AC.
At that point, ∠BAC=60°and BC=75m. Let AB=x meters.
Presently AB/BC=coses60°=2/ √3
=> x/75=2/√3 =>x=150/√3 =150* √3/3=50 √3m.
∴ Length of the string=50 √3m.

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)

A - 5.78m

B - 6.83m

C - 7.24m

D - 6.7m

Answer : B


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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