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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 - 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=>162- X2=162/2=>X2=128 =>X=8√2 =>Height of pole BC = X+4X=5X=40√2
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 - The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.
Answer : B
Explanation
Let AB be the building and AC be its shadow. Then, AC=20m and ∠ACB=60°.Let AB= x m. Presently AB/AC=tan 60°=√3=>x/10=√3 =>x=10√3m= (10*1.732) m=17.32m. ∴ Height of the building is 17.32m.
Q 5 - From The highest point of a 10 m high building, the edge of rise of the of the highest point of a tower is 60° and the despondency's edge of its foot is 45°,Find The tower's stature. (take√3=1.732)
Answer : D
Explanation
Let AB be the building and CD be the tower. Draw BE perpendicular to CD. At that point CE =AB = 10m, ∠EBD= 60° and ∠ACB= ∠ CBE=45° AC/AB= cot45°=1 = >AC/10 =1 => AC = 10m. From △ EBD, we have DE/BE= tan 60°=√3 => DE/AC= √3 => DE/10= 1.732 =>DE = 17.3 Height of the tower = CD= CE+DE= (10+17.32) = 27.3 m.
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 - From a point on a scaffold over the waterway, the edge of dejection of the banks on inverse sides of the waterway is 30°and 45°respectively. In the event that the scaffold is at tallness of 2.5m from the banks, find the width of the Stream. (Take √3=1.732)
Answer : B
Explanation
Let and B be two point on the banks on inverse sides of the stream. Let P be a point on the scaffold at stature of 2.5m. Let PQ-AB. PQ=2.5m.∠BAP=30°and ∠ABP=45°. QB/PQ=cot45°=1 => QB/2.5=1 => QB=2.5m. AQ/PQ =cot30°=√3 => AQ/2.5= √3 => AQ= (2.5)√3m. Width of the stream =AB= (AQ+QB)=2.5(√3+1) 5/2(1.732+1) m=6.83m.