**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)tan45°=ABBC=>1=AB100=> AB = 100 ⋯(1) tan30°=ABBDtan30°=ABBD 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. Required speed =DistanceTime=DistanceTime |

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)tan60°=ACCE=>3=15CE=>CE = 153⋯(1)

tan30°=ABBD=>1√3=15−hBDtan30°=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)tan30°=DCAC=>13=hAC=>AC = h3⋯(1) tan45°=DCBC=>1=hBC=>BC=h⋯(2)tan45°=DCBC=>1=hBC=>BC=h⋯(2) We know that, AB = (AC – BC) =>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. 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 |

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 mtan30°=ACBC=>tan30°=80BC=>BC = 80tan30°=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)tan45°=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)tan60°=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)tan60°=ACCE=>3=100CE=>CE = 1003⋯(1)

tan30°=ABBD=>1√3=100−hBDtan30°=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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