# 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?

A - 10 min 30 sec

B - 13 min 20 sec

C - 15 min 10 sec

D - 16 min 30 sec

### Explanation ```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=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 - 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)

### Explanation ```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
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 - 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

### Explanation ```From right angled triangle ADB,
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?

A - 4.23 m

B - 5.23 m

C - 6.23 m

D - 7.23 m

### Explanation ```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)

A - 135.5m

B - 136.5 m

C - 137.5 m

D - 138.5m

### Explanation ```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
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 - 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°

### Explanation ```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 - 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

A - 15 m

B - 14.86 m

C - 15.64 m

D - 15.8 m

### Explanation ```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 - 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)

A - 4.6m

B - 4.8m

C - 5.2m

D - 5.4m

### Explanation ```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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