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 observes the elevation of a balloon to be 45° at a point A .He then walks towards the balloon and at a certain place B finds the elevation to be 60°. He further walks in the direction of the balloon and finds it to be directly over him at a height of 450 m. Distance travelled from A to B is
Answer : A
Explanation
450/BD= tan (60) =>BD =450/√3 450/AD= tan (30) =>AD= 450√3 AD =BD +AB =>AB=AD-BD= 450√3-450/√3=(450x3-450)/√3=300√3m
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?
Answer : C
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 From the right △ ADB, 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 - 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 - 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 - 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 - 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.
