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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 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 - 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 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
Explanation
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 - 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 - 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)
Answer : B
Explanation
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 - A 10 m long stepping stool is put against a divider. It is slanted at a point of 30°to the ground. The separation of the stepping stool's foot from the divider is:
Answer : C
Explanation
Let AB be the step slanted at 30°to the Ground AC. Then, AB=10m and
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 - On the level plane, there is a vertical tower with a flagpole on its top. At a point 9m far from the tower, the edges of rise of the top and Base of the flagpole are 60°and 30°respectively.The flagpole's tallness is:
Answer : A
Explanation
Let AB be the tower and BC be the flag pole and let O be the point of observation. Then, A=9m, ∠AOB=30°and ∠AOC=60° AB/OA=tan30°=1 ∠ =>AB/9=1∠ =>AB=(9*1/ √3* √3/√3)= 3 √3m. AC/AO=tan60°=√3 =>AC/9= √3 =>AC= 9√3m. ∴BC= (AC-AB) = (9 √3-3 √3) m=6 √3m. ∴ Height of the flagpole is 6 √m.
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