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Problems on Height and Distance – Solved Examples

Problems on Height and Distance – Solved Examples

8. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?
A. 26.28 km/hr B. 32.42 km/hr
C. 24.22 km/hr D. 31.25 km/hr

 

answer with explanation

Answer: Option A

Explanation:

Consider the diagram shown above.

Let AB be the tower. Let C and D be the positions of the boat

Then, ACB = 45° , ADC = 30°, BC = 100 m

tan45°=ABBC=>1=AB100=> AB = 100 ⋯(1)tan⁡45°=ABBC=>1=AB100=> AB = 100 ⋯(1)

tan30°=ABBDtan⁡30°=ABBD
=>1√3=100BD=>13=100BD(∵ Substituted the value of AB from equation 1)
=> BD =100√3=1003

CD = (BD – BC) =(100√3−100)=100(√3−1)=(1003−100)=100(3−1)

It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√3−1)100(3−1) is covered in 10 seconds.

Required speed =DistanceTime=DistanceTime
=100(√3−1)10=10(1.73−1)=100(3−1)10=10(1.73−1)
= 7.3 meter/seconds
= 7.3 × 185185 km/hr = 26.28 km/hr

 

9. The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 5 metres B. 8 metres
C. 10 metres D. 12 metres

answer with explanation

Answer: Option C

Explanation:

Consider the diagram shown above. AC represents the tower and DE represents the pole.

Given that AC = 15 m, ADB = 30°, AEC = 60°

Let DE = h
Then, BC = DE = h,
AB = (15-h)    (∵ AC=15 and BC = h),
BD = CE

tan60°=ACCE=>√3=15CE=>CE = 15√3⋯(1)tan⁡60°=ACCE=>3=15CE=>CE = 153⋯(1)

tan30°=ABBD=>1√3=15−hBDtan⁡30°=ABBD=>13=15−hBD
=>1√3=15−h(15√3)=>13=15−h(153)(∵ BD = CE and substituted the value of CE from equation 1 )
=>(15−h)=1√3×15√3=153=5=>h=15−5=10 m=>(15−h)=13×153=153=5=>h=15−5=10 m

i.e., height of the electric pole = 10 m

10. The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:
A. 64.2 m B. 62.2 m
C. 52.2 m D. 54.6 m

 

answer with explanation

Answer: Option D

Explanation:

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m

We have DAC = 30°, DBC = 45°

Let DC = h

tan30°=DCAC=>1√3=hAC=>AC = h√3⋯(1)tan⁡30°=DCAC=>13=hAC=>AC = h3⋯(1)

tan45°=DCBC=>1=hBC=>BC=h⋯(2)tan⁡45°=DCBC=>1=hBC=>BC=h⋯(2)

We know that, AB = (AC – BC)
=> 40 = (AC – BC)
=> 40=(h√3−h)40=(h3−h)[∵ from (1) & (2)] =>40=h(√3−1)

=>h=40(√3−1)=40(√3−1)×(√3+1)(√3+1)=

40(√3+1)(3−1)=40(√3+1)2

=20(√3+1)=20(1.73+1)

=20×2.73=54.6 m

 

11. On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m, the distance between the objects is approximately equal to :
A. 272 m B. 284 m
C. 288 m D. 254 m

 

answer with explanation

Answer: Option D

Explanation:

Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m , DAC = 45°, DBC = 60°

tan 60°=DCBC√3=600BCBC= 600√3⋯(1)tan 60°=DCBC3=600BCBC= 6003⋯(1)

tan 45°=DCAC1=600ACAC= 600⋯(2)tan 45°=DCAC1=600ACAC= 600⋯(2)

Distance between the objects
= AB = (AC – BC)
=600−600√3=600−6003 [∵ from (1) and (2)] =600(1−1√3)=600(√3−1√3)=600(√3−1√3)×√3√3=600√3(√3−1)3=200√3(√3−1)=200(3−√3)=200(3−1.73)=254 m=600(1−13)=600(3−13)=600(3−13)×33=6003(3−1)3=2003(3−1)=200(3−3)=200(3−1.73)=254 m

 

12. A ladder 10 m long just reaches the top of a wall and makes an angle of 60° with the wall.Find the distance of the foot of the ladder from the wall (√3=1.73)(3=1.73)
A. 4.32 m B. 17.3 m
C. 5 m D. 8.65 m

answer with explanation

Answer: Option D

Explanation:

Let BA be the ladder and AC be the wall as shown above.
Then the distance of the foot of the ladder from the wall = BC

Given that BA = 10 m , BAC = 60°

sin 60°=BCBA√32=BC10BC = 10×√32=5×1.73=8.65 msin 60°=BCBA32=BC10BC = 10×32=5×1.73=8.65 m

13. From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
A. 40 m B. 138.4 m
C. 46.24 m D. 160 m

answer with explanation

Answer: Option B

Explanation:

Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30°

ABC = DAB = 30°   (because DA || BC)

tan30°=ACBC=>tan30°=80BC=>BC = 80tan30°=80(1√3)=80×1.73=138.4 mtan⁡30°=ACBC=>tan⁡30°=80BC=>BC = 80tan⁡30°=80(13)=80×1.73=138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m

14. The angle of elevation of the top of a lighthouse 60 m high, from two points on the ground on its opposite sides are 45° and 60°. What is the distance between these two points?
A. 45 m B. 30 m
C. 103.8 m D. 94.6 m

answer with explanation

Answer: Option D

Explanation:

Let BD be the lighthouse and A and C be the two points on ground.
Then, BD, the height of the lighthouse = 60 m

BAD = 45° ,  BCD = 60°

tan45°=BDBA⇒1=60BA⇒BA=60 m⋯(1)tan⁡45°=BDBA⇒1=60BA⇒BA=60 m⋯(1)

tan60°=BDBC⇒√3=60BC⇒BC=60√3=60×√3√3×√3=60√33=20√3=20×1.73=34.6 m⋯(2)tan⁡60°=BDBC⇒3=60BC⇒BC=603=60×33×3=6033=203=20×1.73=34.6 m⋯(2)

Distance between the two points A and C
= AC = BA + BC
= 60 + 34.6   [∵ Substituted value of BA and BC from (1) and (2)] = 94.6 m

15. From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
A. 52 m B. 50 m
C. 66.67 m D. 33.33 m

answer with explanation

Answer: Option C

Explanation:

Consider the diagram shown above. AC represents the hill and DE represents the pole

Given that AC = 100 m
XAD = ADB = 30°   (∵ AX || BD )
XAE = AEC = 60°   (∵ AX || CE)

Let DE = h

Then, BC = DE = h,
AB = (100-h)   (∵ AC=100 and BC = h),
BD = CE

tan60°=ACCE=>√3=100CE=>CE = 100√3⋯(1)tan⁡60°=ACCE=>3=100CE=>CE = 1003⋯(1)

tan30°=ABBD=>1√3=100−hBDtan⁡30°=ABBD=>13=100−hBD
=>1√3=100−h(100√3)=>13=100−h(1003)(∵ BD = CE and substituted the value of CE from equation 1)
=>(100−h)=1√3×100√3=1003=33.33

=>h=100−33.33=66.67 m

=> (100−h)=13×1003=1003=33.33

=>h=100−33.33=66.67 m

i.e., the height of the pole = 66.67 m

 

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