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?
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 - 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?
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 - 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
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 - 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
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 - Two men are inverse sides of a tower. They gauge the edge of the rise of the highest point of the tower as 30° and 45° respectively. On the off chance that the tallness of the tower is 50 m, discover the separation between the two men. (Take √3=1.732)
Let AB be the tower and let C and D be the two's positions men.
At that point ∠ACB=30°,∠ADB= 45°and AB= 50 m
AC/AB = Cot30°=√3 => AC/50 = √3
=>AC=50√3m
AD/AB=cot 45°=1 => AD/50=1
=> AD=50M.
Separation between the two men =CD= (AC+AD)
= (50√3+50) m=50(√3+1)
=50(1.73+1)m=(50*2.73)m=136.5m.
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:
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
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)
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.
