# 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 - A vertical pole fixed to the ground is divided in the ratio 1:4 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 16 m away from the base of the pole, what is the height of the pole?

### Answer : C

### Explanation

Let CB be the pole and point D divides it such that BD : DC = 1 : 4 = X:4X Given that AB = 16 m Let the the two parts subtend equal angles at point A such that CAD = BAD = Θ =>tan Θ=X/16 =>X=16 tan ( Θ) ------ (1) =>tan( Θ+ Θ)=4X/16 =>16 tan (2 Θ)=4X =>16(2tan ( Θ))/(1-tan ( Θ)^{2})=4X ------ (2) From eqn 1 & 2 2X/(1-tan ( Θ)^{2})=4X (X=16tan Θ) 1/(1-(X/16)^{2})=2 1-(X/16)^{2}=1/2=>16^{2}- X^{2}=162/2=>X^{2}=128 =>X=8√2 =>Height of pole BC = X+4X=5X=40√2

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

### Answer : A

### Explanation

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?

### Answer : A

### 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 - Consider vertical shaft 6 m high has a shadow of length 2√3m, discover the angle of elevation of the sun?

### Answer : A

### Explanation

Let AB be the building and AC be its shadow. Then, AB= 6m and AC= 2√3m. Let ∠ACB= θ Then tan θ = AB/AC= 6/2√3m =√3= >θ =60° Point of rise of the sun is 60°

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:

### Answer : A

### 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 - 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

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

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.