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Problems on Height and Distance – Solved Examples

Problems on Height and Distance – Solved Examples

1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 300 m B. 173 m
C. 273 m D. 200 m

answer with explanation

Answer: Option C

Explanation:

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,  BAD = 30° ,  BCD = 45°

tan30°=BDBA⇒1√3=100BA⇒BA=100√3tan⁡30°=BDBA⇒13=100BA⇒BA=1003

tan45°=BDBC⇒1=100BC⇒BC=100tan⁡45°=BDBC⇒1=100BC⇒BC=100

Distance between the two ships
= AC = BA + BC
=100√3+100=100(√3+1)=1003+100=100(3+1)
= 100(1.73+1) = 100 × 2.73 = 273 m

2. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 45º. What is the distance between the base of the tower and the point P?
A. 9 units B. 3√333 units
C. Data inadequate D. 12 units

 

 

answer with explanation

Answer: Option C

Explanation:

tan 45°=SRQRtan 30°=SRPR=SR(PQ + QR)tan 45°=SRQRtan 30°=SRPR=SR(PQ + QR)

Two equations and 3 variables. Hence we can not find the required value with the given data.

(Note that if one of SR, PQ, QR is known, this becomes two equations and two variables and if that was the case, we could have found out the required value.)

 

3. From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
A. 346 m B. 400 m
C. 312 m D. 298 m

 

answer with explanation

Answer: Option A

Explanation:

tan30°=RQPQ1√3=200PQPQ=200√3=200×1.73=346 mtan⁡30°=RQPQ13=200

PQPQ=2003=200×1.73=346 m

 

4. The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
A. None of these B. 60°
C. 45° D. 30°

answer with explanation

Answer: Option C

Explanation:

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let QPR = θ

tanθ=QRPQ=1  tan⁡θ=QRPQ=1   (since QR = PQ)
=> θ = 45°

i.e., required angle of elevation = 45°

5. An observer 2 m tall is 10√3103 m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. None of these B. 12 m
C. 14 m D. 10 m

 

answer with explanation

Answer: Option B

Explanation:

SR = PQ = 2 m
PS = QR =10√3=103 m

tan30°=TSPS1√3=TS10√3TS=10√3√3=10 mtan⁡30°=TSPS13=TS103TS=1033=10 m

TR = TS + SR = 10 + 2 = 12 m

 

6. The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
A. 14.8 m B. 6.2 m
C. 12.4 m D. 24.8 m

answer with explanation

Answer: Option D

Explanation:

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60° = PQPR12=12.4PRPR=2×12.4=24.8 mcos 60° = PQPR12=12.4PRPR=2×12.4=24.8 m

7. A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
A. 8 min 17 second B. 10 min 57 second
C. 14 min 34 second D. 12 min 23 second

 

answer with explanation

Answer: Option B

Explanation:

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.

Then, ADC = 30° , ACB = 45°

Let AB = h, BC = x, CD = y

tan45°=ABBC=hx=>1=hx=>h=x⋯(1)tan⁡45°=ABBC=hx=>1=hx=>h=x⋯(1)

tan30°=ABBD=AB(BC + CD)=hx+y=>1√3=hx+y=>x + y = √3h=>y = √3h – xtan⁡30°=ABBD=AB(BC + CD)

=hx+y

=>13=hx+y

=>x + y = 3h

=>y = 3h – x
=>y = √3h−h  =>y = 3h−h  (∵ Substituted the value of x from equation 1 )
=>y = h(√3−1)=>y = h(3−1)

Given that distance y is covered in 8 minutes.
i.e, distance h(√3−1)h(3−1) is covered in 8 minutes.

Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes.

since distance is proportional to the time when the speed is constant, we have

h(√3−1)∝8⋯(A)h∝t⋯(B)(A)(B)=>h(√3−1)h=8t=>(√3−1)=8t

=>t=8(√3−1)=8(1.73−1)=8.73=80073

minutes=107073minutes≈10 minutes 57 secondsh(3−1)∝8⋯(A)h∝t⋯(B)(A)(B)=>h(3−1)h=8t

=>(3−1)=8t=>t=8(3−1)=8(1.73−1)=8.73=80073minutes=107073minutes≈10 minutes 57 seconds

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