10th Maths 9.1

Chapter 9 

Some Applications of Trigonometry

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NCERT Class 10th solution of Exercise 8.1

NCERT Class 10th solution of Exercise 10.1

8th, 9th, 

Exercise 9.1

Q1. A circus artist is climbing a `20m` long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is `30^circ`(see Figure).

`text{Sol.}`

`text{Let the height of the pole is h.}`

`text{In right }triangle text{ABC}`

`text{sin }30^circ=text{AB}/text{BC}`

`1/2=text{h}/20`

`text{h }=20/2`

`text{h = 10 m}`

`text{Answer:}`

`text{The height of the pole is 10 m.}`

Q2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle `30^circ` with it. The distance between the foot of the tree to the point where the top touches the ground is `8m`. Find the height of the tree.

`text{Sol.}`

`text{Let the height of tree x + y}`

`text{In right }triangle text{ABC}`

`text{tan }30^circ = text{AB}/text{BC}`

`1/sqrt3 = text{x}/8`

`text{x}=8/sqrt3`_________(1)

`text{cos }30^circ =text{BC}/text{AC}`

`sqrt3/2 = 8/text{y}`

`text{y}=(8times2)/sqrt3`

`text{y}=16/sqrt3`________(2)

`text{x+y}=8/sqrt3+16/sqrt3``text{ [from (1) and (2)]}`

`text{x+y}=(8+16)/sqrt3`

`text{x+y}=24/sqrt3`

`text{x+y}=(24timessqrt3)/(sqrt3timessqrt3)``text{  [Rationalisation]}`

`text{x+y}=(24timessqrt3)/3`

`text{x+y}=8sqrt3`

`text{Answer:}`

`text{The height of the tree is 8}sqrt3text{ m}`.

Q3. A contractor plans to install two slides for the children to play in a park. For the children below the age of `5` years, she prefers to have a slide whose top is at a height of `1.5 m`, and is inclined at an angle of `30^circ` to the ground, whereas for elder children, she wants to have a steep slide at a height of `3m`, and inclined at an angle of `60^circ` to the ground. What should be the length of the slide in each case?

`text{Sol.}`

`text{first case}`

`text{Let the length of the slide x}`

`text{In right } triangletext{ABC}`

`text{sin }30^circ=text{BC}/text{AC}`

`1/2=1.5/text{x}`

`text{x}=1.5times2`

`text{x}=3text{ m}`

`text{Second Case}`

`text{Let the length  of the slide is y}`

`text{In right }triangletext{PQR}`

`text{sin }60^circ=text{QR}/text{PR}`

`sqrt3/2=3/text{y}`

`text{y}=(3times2)/sqrt3`

`text{y}=(3times2timessqrt3)/(sqrt3timessqrt3)``text{  [Rationalisation]}`

`text{y}=(3times2timessqrt3)/3`

`text{y}=2sqrt3text{ m}`

`text{Answer:}`

`text{The length of slide in first case 3 m }``text{and second case 2}sqrt3 text{ m}`.

Q4. The angle of elevation of the top of a tower from a point on the ground, which is `30m` away from the foot of the tower, is `30^circ`. Find the height of the tower.

`text{Sol.}`

`text{Let the height of the tower is h}`

`text{In right }triangletext{ABC}`

`text{tan }30^circ=text{AB}/text{BC}`

`1/sqrt3=text{h}/30`

`text{h}=30/sqrt3`

`text{h}=(30timessqrt3)/(sqrt3timessqrt3)``text{  [Rationalisation]}`

`text{h}=(30timessqrt3)/3`

`text{h}=10timessqrt3text{ m}`

`text{Answer:}`

`text{The height of tower is 10}sqrt3 text{ m}.`

Q5. A kite is flying at a height of `60m` above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is `60^circ`. Find the length of the string, assuming that there is no slack in the string.

`text{Sol.}`

`text{Let the length of the string is x}`

`text{In right }triangle text{ABC}`

`text{sin }60^circ=text{BC}/text{AC}`

`sqrt3/2=60/text{x}`

`text{x}=(60times2)/sqrt3`

`text{x}=(60times2timessqrt3)/(sqrt3timessqrt3)``text{  [Rationalisation]}`

`text{x}=(60times2timessqrt3)/3`

`text{x}=20times2timessqrt3`

`text{x}=40sqrt3text{ m}`

`text{Answer:}`

`text{The length of the string is 40}sqrt3text{ m}`.

Q6. A `1.5m` tall boy is standing at some distance from a `30m` tall building. The angle of elevation from his eyes to the top of the building increases from `30^circ` to `60^circ` as he walks towards the building. Find the distance he walked towards the building.

`text{Sol.}`

`text{Let the distance towards the building DA = x, AB = y}`

`text{In right }triangletext{ABC}`

`text{tan }60^circ=text{AB}/text{BC}`

`sqrt3=28.5/text{BC}`

`text{BC}=28.5/sqrt3`

`text{BC}=(28.5timessqrt3)/(sqrt3timessqrt3)``text{ [Rationalisation]}`

`text{BC}=9.5sqrt3`______(1)

`text{In right }triangletext{ABD}`

`text{tan }30^circ=text{AB}/text{DB}`

`1/sqrt3=28.5/(text{DC + BC})``text{ [from (1)]}`

`text{DC + BC} = 28.5sqrt3`

`text{DC + }9.5sqrt3 = 28.5sqrt3`

`text{DC }= 28.5sqrt3-9.5sqrt3`

`text{DC }=(28.5 - 9.5)sqrt3`

`text{DC }=19sqrt3 text{ m}`

`text{Answer:}`

`text{The distance towards the building 19}sqrt3text{ m}.`

Q7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a `20m` high building are `45^circ` and `60^circ` respectively. Find the height of the tower.

`text{Sol.}`

`text{Let the of the tower is h}`

`text{In right } triangle text{ABC}`

`text{tan }45^circ=text{AB}/text{BC}`

`1=20/text{BC}`

`text{BC}=20 text{ m}`_______(1)

`text{In right } triangletext{DBC}`

`text{tan }60^circ=text{BD}/text{BC}`

`sqrt3=(20+text{h})/text{BC}`

`sqrt3=(20+text{h})/20``text{  [from (1)]}`

`20sqrt3=20+text{h}`

`text{h}=20sqrt3-20`

`text{h}=20(sqrt3-1)`

`text{Answer:}`

`text{The height of the tower is }20(sqrt3-1) text{ m}.`

Q8. A statue, `1.6m` tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is `60^circ` and from the same point the angle of elevation of the top of the pedestal is `45^circ`. Find the height of the pedestal.

`text{Sol.}`

`text{Let the height of the pedestal is h.}`

`text{In right } triangle text{ABC}`

`text{tan }45^circ=text{AB}/text{BC}`

`1=text{h}/text{BC}`

`text{h }=text{AB}`

`text{In right } triangle text{DBC}`

`text{tan }60^circ=text{BD}/text{BC}`

`sqrt3=(text{h}+1.6)/text{h}`

`text{h }sqrt3=text{h}+1.6`

`text{h }sqrt3-text{h}= 1.6`

`text{h }(sqrt3-1)= 1.6`

`text{h }=1.6/(sqrt3-1)`

`text{h }=(1.6times(sqrt3+1))/((sqrt3-1)times(sqrt3+1))``text{  [Rationalisation]}`

`text{h }=(1.6times(sqrt3+1))/(3-1)`

`text{h }=1.6times(sqrt3+1)/2`

`text{h }=0.2times(sqrt3+1)text{ m}`

`text{Answer:}`

`text{The height of the pedestal is }0.2(sqrt3+1)text{ m}`.

Q9. The angle of elevation of the top of a building from the foot of the tower is `30^circ` and the angle of elevation of the top of the tower from the foot of the building is `60^circ`. If the tower is `50m` high, find the height of the building.

`text{Sol.}`

`text{Let the height of the building is h.}`

`text{In right }triangle text{ABC}`

`text{tan }30^circ=text{AB}/text{BC}`

`1/sqrt3=text{h}/text{BC}`

`text{BC }=text{h}sqrt3`______(1)

`text{In right }triangle text{BCD}`

`text{tan }60^circ=text{CD}/text{BC}`

`sqrt3=50/(text{h}sqrt3)``text{ [from (1)]}`

`text{h}sqrt3times sqrt3=50`

`3text{h }=50`

`text{h }=50/3`

`text{h }=16(2)/3text{ m}`

`text{Answer:}`

`text{The height of the building is }16(2)/3text{ m}.`

Q10. Two poles of equal heights are standing opposite each other on either side of the road, which is `80m` wide. From a point between them on the road, the angles of elevation of the top of the poles are `60^circ` and `30^circ`, respectively. Find the height of the poles and the distance of the point from the poles.

`text{Sol.}`

`text{Let the height of the poles are h.}`

`text{and the BC = x m, CQ = (80-x)m}` 

`text{In right }triangle text{ABC}`

`text{tan }30^circ=text{AB}/text{BC}`

`1/sqrt3=text{h}/text{x}`

`text{x }=text{h}sqrt3`_______(1)

`text{In right }triangletext{CQR}`

`text{tan }60^circ=text{QR}/text{CQ}`

`sqrt3=text{h}/text{80-x}`

`text{h }=sqrt3(80-text{h}sqrt3)``text{ [From (1)]}`

`text{h }=80sqrt3-3text{h}`

`text{h+3h }=80sqrt3`

`text{4h }=80sqrt3`

`text{h }=(80sqrt3)/4`

`text{h }=20sqrt3text{ m}`

`text{This value put in eq. (1)}`

`text{x  }=20sqrt3 times sqrt3`

`text{x  }=20 times 3`

`text{x }=60 text{ m}`

`text{and }`

`text{CQ = 80 - x} = 80 - 60 = 20 text{ m}` 

`text{Answer:}`

`text{The height of the poles are }20sqrt3text{ m. and }``text{the distance of the point from the poles are}``text{60 m and 20 m}.`

Q11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is `60^circ`. From another point `20m` away from this point in the line joining this point to the tower, the angle of elevation of the top of the tower is `30^circ` (see Figure). Find the height of the tower and the width of the canal.

`text{Sol.}`

`text{Let the height of the tower is h and the width is x}`

`text{In right }triangle text{ABC}`

`text{tan }30^circ=text{AB}/text{BC}`

`1/sqrt3=text{h}/(text{CD+DB})`

`1/sqrt3=text{h}/(text{20+DB})`

`text{20+DB }=text{h}sqrt3`

`text{DB }=text{h}sqrt3-20`_______(1)

`text{In right }triangle text{DBC}`

`text{tan }60^circ=text{BC}/text{DB}`

`sqrt3=text{h}/(text{h}sqrt3-20)``text{ [from (1)]}`

`text{h }=sqrt3(text{h}sqrt3-20)`

`text{h }=3text{h}-20sqrt3`

`text{3h-h }=20sqrt3`

`text{2h }=20sqrt3`

`text{h }=(20sqrt3)/2`

`text{h }=10sqrt3`

`text{Put in eq.(1)}`

`text{DB }=text{10}sqrt3times sqrt3-20`

`text{DB }=10times3-20`

`text{DB }=30-20`

`text{DB }=10text{ m}`

`text{Answer:}`

`text{The height of the tower is }10sqrt3text{ m }``text{and the width is }10text{ m}.`

Q12. From the top of a `7m` high building, the angle of elevation of the top of a cable tower is `60^circ` and the angle of depression of its foot is `45^circ`. Determine the height of the tower.

`text{Sol.}`

`text{Let the height of the tower is h}.`

`text{In right }triangle text{ABC}`

`text{tan }45^circ=text{AB}/text{BC}`

`1=7/text{BC}`

`text{BC }=7`_______(1)

`text{In right }triangle text{PBC}`

`text{tan }60^circ=text{BP}/text{BC}`

`sqrt3=text{h}/7``[text{From eq (1)}]`

`text{h }=7sqrt3`

`text{PC }= text{ h+7}`

`text{PC }=7sqrt3+7`

`text{PC }=7(sqrt3 + 1)text{ m}`

`text{Answer:}` 

`text{The height of the tower is }7(sqrt3+1)text{ m}.`

Q13. As observed from the top of a `75m` high lighthouse from the sea level, the angles of depression of two ships are `30^circ` and `45^circ`. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

`text{Sol.}`

`text{Let the distance between the two ships x.}`

`text{In right }triangle text{ABD}`

`text{tan }45^circ=(text{AB})/(text{DB})`

`1=75/(text{y})`

`text{y}=75`______(1)

`text{In right }triangle text{ABC}`

`text{tan }30^circ=(text{AB})/(text{BC})`

`1/sqrt3=75/(text{x+y})`

`1/sqrt3=75/(text{x+75})``[text{ From eq. (1)}]`

`text{x+75}=75sqrt3text{ m}.`

`text{x}=75sqrt3-75`

`text{x}=75(sqrt3-1)``text{ m}`

`text{Answer:}`

`text{The distance between the two ships }``75(sqrt3-1)text{ m}`.

Q14. A `1.2m` tall girl spots a balloon moving with the wind in a horizontal line at a height of `88.2m` from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is `60^circ`. After some time, the angle of elevation reduces to `30^circ` (see Figure). Find the distance traveled by the balloon during the interval.

`text{Sol.}`

`text{Let the distance of the balloons x.}`

`text{In right }triangletext{ABC}`

`text{tan}60^circ=(text{AB})/(text{BC})`

`sqrt3=87/(text{BC})`

`text{BC}=87/sqrt3`

`text{In right }triangle text{PQC}`

`text{tan}30^circ=(text{PQ})/(text{QC})`

`1/sqrt3=87/(text{BC+BQ})`

`text{BC+BQ}=87sqrt3`

`87/sqrt3 + text{BQ}=87sqrt3``[text{ From eq. (1)}]`

`text{BQ}=87sqrt3-87/sqrt3`

`text{BQ}=(87times3-87)/sqrt3`

`text{BQ}=(87(3-1))/sqrt3`

`text{BQ}=(87times2)/sqrt3`

`text{BQ}=(174timessqrt3)/(sqrt3timessqrt3)``text{  [Rationalisation]}`

`text{BQ}=(174timessqrt3)/3`

`text{BQ}=58sqrt3text{ m}`

`text{Answer:}`

`text{The distance of the balloons 58}sqrt3text{ m}.`

Q15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of `30^circ`, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be `60^circ`. Find the time taken by the car to reach the foot of the tower from this point.

`text{Sol.}`

`text{Let car cover the distance in x seconds}`

`text{In right }triangle text{DBC}`

`text{tan }60^circ=text{BC}/text{DB}`

`sqrt3=text{h}/text{tx}`

`sqrt3text{ tx}=text{h}`_______(1)

`text{In right }triangletext{ABC}`

`text{tan }30^circ=text{BC}/text{AB}`

`1/sqrt3=text{h}/(text{AD+DB})`

`1/sqrt3=(text{tx}sqrt3)/(6x+text{tx})``[text{ From (1)}]`

`6x+text{tx}=text{tx}sqrt3times sqrt3`

`text{x(6+t)}=3text{tx}`

`text{3t-t} = 6`

`text{2t} = 6`

`text{t}=6/2`

`text{t}=3text{ seconds}`

`text{Answer:}`

`text{The car cover the distance in 3 seconds}`

Q16. The angles of elevation of the top of a tower from two points at a distance of `4m` and `9m` from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is `6m`.

`text{Sol.}`

`text{Let the height of the tower is h}`

`text{In right }triangle text{ PQR}`

`text{tan }theta=text{PQ}/text{AQ}`

`text{tan }theta=text{h}/4`

`text{h }=4text{ tan }theta`_______(1)

`text{In right }triangle text{ PQB}`

`text{tan }(90 - theta)=text{PQ}/text{BQ}`

`text{cot }theta=text{h}/9`_______(2)

`text{h}timestext{h}=9text{ tan }theta times4text{ cot }theta``text{ [from (1) and (2)]}`

`text{h}^2=36`

`text{h}=sqrt(36)`

`text{h}=6text{ m}`

`text{Answer:}`

`text{The height of the tower is 6 m.}`