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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 - When the sun's altitude changes from 45° to 60°, the length of the shadow of a tower decreases by 45m. What is the height of the tower?
Answer : A
Explanation
Let AD be the tower, BD be the initial shadow and CD be the final shadow. Given that BC = 45 m, ABD = 45°, ACD = 60°, Let CD = x, AD = h From the right CDA, tan60=h/x From the right BDA, tan45=(45+x)/h=>h=45+x =>h=45+h/√3 =>h(1-1/√3)=45 =>h=45/(1-1/√3)=(45√3)/(√3-1)
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 - 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
Explanation
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
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 - From The highest point of a bluff 90 m high, the edges of Misery of the top and base of a tower are seen to be 30° and 60°. What is the tower's tallness is?
Answer : C
Explanation
Let AB be the precipice and CD be the tower. Draw DE || CA. Then, ∠BDE=30°, ∠BCA=60°and AB= 90m. From right △CAB, we have CA/AB=cost60°=1/√3 => CA/90=1/√3 =>CA=(90*1/√3* √3/√3) =30 √3m. ∴ DE =CE=30/√3m. From right ?DEB, we have BE/DE= tan30°=1/√3 => BE/30 √3=1 √3 =>BE= (30 √3*1 √3) =30m. ∴ CD=AE= (AB-BE) = (90-30) m=60m. Hence, the tower's stature is 60m.
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
Explanation
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.