# 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 - The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

### Answer : D

### Explanation

Let AB be the wall and BC be the ladder. Then, ∠ACB = 45° and AC = 7.5 m AC/BC= Cos (45) =1/√2 BC=7.5√2

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 - 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

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

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 - 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 - 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

### Answer : A

### 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)

### 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.