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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 on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 10 minutes for the angle of depression to change from 45° to 60°, how soon after this will the car reach the observation tower?

Answer : B

Explanation

Height & Distance Solution 3

Let AB be the tower and C and D be the two positions of the car.
Then,from figure
AB/AC=tan 60 =√3 => AB=√3AC
AB/AD=tan 45=1 => AB=AD
AB=AC+CD
CD=AB-AC=√3AC - AC=AC (√3-1)
CD = AC (√3-1) =>10 min
 AC=> ?
AC/(AC(√3-1)) x 10=?=10/(√3-1)=13.66=13 min 20 sec(approx)

Q 2 - When the sun's altitude changes from 45° to 60°, the length of the shadow of a tower decreases by 45m. What is the height of the tower?

Answer : A

Explanation

Height & Distance Solution 6

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 45 m,  ABD = 45°,  ACD = 60°,
Let CD = x, AD = h
From the right   CDA, tan60=h/x
From the right   BDA, tan45=(45+x)/h=>h=45+x
=>h=45+h/√3
=>h(1-1/√3)=45
=>h=45/(1-1/√3)=(45√3)/(√3-1)

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

Answer : C

Explanation

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)
=>CD+180=180√3
=>CD=180(√3-1)
Speed =Distance/Time=180(√3-1)/90=2(√3-1) m/sec

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?

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 - 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)

Answer : B

Explanation

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
=>AC=50√3m
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)
=50(1.73+1)m=(50*2.73)m=136.5m.

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 - 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?

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 stature of a tree is 10m. It is twisted by the wind in a manner that its top touches the ground and makes a point of 60 with the ground. At what range from base did the tree get twisted? (√3=1.73)

Answer : A

Explanation

Height & Distance Solution 23

Let AB be the tree bowed at the point C so that part CB takes the position CD.
Then ,CD=CB. Let AC=x meters. At that point, CD = CB= (10-X) m and ∠ADC=60°.
AC/AC=sin60° => x/(10-x) = √3/2
=>2x=10 √3-√3x
=> (2+ √3) x= 10 √3
=>x=10 √3/ (2+ √3)*(2-√3)/(2-√3)=20 √3-30)m
= (20*1.73-30) m=4.6m
=> Required height=4.6m.

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