Aptitude - Height & Distance Online Quiz


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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 - 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

A - 300√3 m

B - 200√3 m

C - 100√3 m

D - 450√3 m

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 - 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

Explanation

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,
Tan45=CB/AB=>CB=AB
From the right △ ADB,
Tan30=DB/AB=>DB=ABtan30=CBx(1/√3)=750/√3
CB=CD+DB
=> Required height CD=CB-DB=750-750/√3=250(3- √3)

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

A - 210√3

B - 210/√3

C - 70√3

D - 105√3

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 - A step inclining toward a vertical divider makes a point of 45 with the even ground. The step's Foot is 3m from the divider. Find Length of the step?

A - 4.23 m

B - 5.23 m

C - 6.23 m

D - 7.23 m

Answer : A

Explanation

Height & Distance Solution 11

Let AB be the step and BC be the divider and let AC be the even ground.
Then, ∠CAB=45 and AC=3m. Let AB= x Meter.
From right △ ACB, we have
AB/AC =sec. 45° = √2 => x/3 = √2
X= 3√2m = (3*1.41) m= 4.23m.
∴ Length of the stepping stool is 4.23 m

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)

A - 24.3m

B - 25.3m

C - 26.3 m

D - 27.3 m

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 - The point of height of a tower from a separation 50 m from its foot is 30. The tower's tallness is:

A - 50√3m

B - 50/√3m

C - 23√3m

D - 100m/√3

Answer : B

Explanation

Height & Distance Solution 16

Let AB be the tower and AC be the even line such that AC=50 m and ∠ACB=30°.
AB/AC=tan 30°=1/√3
=>x/50 = 1/√3
> x=50*1/√3m= 50/√3m.
∴ Height of the tower=50/√3m.

Q 7 - From The highest point of a bluff 90 m high, the edges of Misery of the top and base of a tower are seen to be 30° and 60°. What is the tower's tallness is?

A - 30 m

B - 45 m

C - 60 m

D - 75 m

Answer : C

Explanation

Height & Distance Solution 21

Let AB be the precipice and CD be the tower. Draw DE || CA.
Then, ∠BDE=30°, ∠BCA=60°and AB= 90m.
From right △CAB, we have
CA/AB=cost60°=1/√3 => CA/90=1/√3
=>CA=(90*1/√3* √3/√3)
=30 √3m.
∴ DE =CE=30/√3m.
From right ?DEB, we have
BE/DE= tan30°=1/√3 => BE/30 √3=1 √3
=>BE= (30 √3*1 √3) =30m.
∴ CD=AE= (AB-BE) = (90-30) m=60m.
Hence, the tower's stature is 60m.

Q 8 - The point of dejection of two boats from the highest point of a beacon are 45°and 30°towards east. On the off chance that the boat are 200 m separated, the light's stature hours is :( Take √3=1.73)

A - 100 m

B - 173 m

C - 200 m

D - 273 m

Answer : D

Explanation

Height & Distance Solution 22

Let AB be the beacon and C and D be the positions the boats such that CD=200m.
∠ABC=45°and ∠ADB=30°.
AB/AC =tan45°=1=>AB=AC=x m.
Presently AB/AD =tan30°=1/√3
h/h+200=1 √3?√3h=h+200
=> (√3-1) h=200
=> h=200/√3-1)*(√3+1/ (√3+1) =100*(√3+1)
=100(1.73+1) =273m.


aptitude_height_distance.htm

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