# 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

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

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 straight tree is broken due to thunder storm. The broken part is bent in such a way that the peak touches the ground at an angle elevation of 45°. The distance of peak of tree (where it touches the root of the tree is 20 m. Then the height of the tree is

### Answer : A

### Explanation

Let the total length of the tree be X+Y meters From the figure tan 45=X/20 =>X=20 cos 45 = 20/Y =>Y=20/cos 45 =20√2 X+Y=20+20radic;2=20+2x10x1.414 =48.28 meters

Q 4 - A flag staff of 10 meters height stands on a building of 50 meters height. An observer at a height of 60 meters subtends equal angles to the flag staff and the building. The distance of the observer from the top of the flag staff is

### Answer : C

### Explanation

From the figure tanθ=10/CB tan(2θ)=60/CB=(2tan(θ))/(1-tan(θ)^{2}) =>60/CB=(2tan(θ))/(1-tan(θ)^{2})=(2(10/CB))/(1-(10/CB)^{2}) =>3/1=1/(1-(10/CB)^{2}) =>3x(1-(10/CB)sup>2)=1 3CB^{2}-300=CB^{2}2CB^{2}=300=>CB=√150=5√6

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

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

### Answer : B

### Explanation

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 - 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:

### Answer : B

### Explanation

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 - 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

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.