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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 - 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
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
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 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
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 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
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 3CB2-300=CB2 2CB2=300=>CB=√150=5√6
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
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
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 - 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
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 point of dejection of two boats from the highest point of a beacon are 45°and 30°towards east. On the off chance that the boat are 200 m separated, the light's stature hours is :( Take √3=1.73)
Answer : D
Let AB be the beacon and C and D be the positions the boats such that CD=200m. ∠ABC=45°and ∠ADB=30°. AB/AC =tan45°=1=>AB=AC=x m. Presently AB/AD =tan30°=1/√3 h/h+200=1 √3?√3h=h+200 => (√3-1) h=200 => h=200/√3-1)*(√3+1/ (√3+1) =100*(√3+1) =100(1.73+1) =273m.