10th Maths 9.1

Chapter 9 Some Applications of Trigonometry

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NCERT Class 10th Maths Projects

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

20 m

$20m$ 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}

$30^circ$ (see Figure).

Sol.

$text{Sol.}$

Let the height of the pole is h.

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

In right △ ABC

$text{In right }triangle text{ABC}$

sin 30^{\circ} = \frac{AB}{BC}

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

\frac{1}{2} = \frac{h}{20}

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

h = \frac{20}{2}

$text{h }=20/2$

h = 10 m

$text{h = 10 m}$

Answer:

$text{Answer:}$

The height of the pole is 10 m.

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

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}

$30^circ$ with it. The distance between the foot of the tree to the point where the top touches the ground is

8 m

$8m$ . Find the height of the tree.

Sol.

$text{Sol.}$

Let the height of tree x + y

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

In right △ ABC

$text{In right }triangle text{ABC}$

tan 30^{\circ} = \frac{AB}{BC}

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

\frac{1}{\sqrt{3}} = \frac{x}{8}

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

x = \frac{8}{\sqrt{3}}

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

cos 30^{\circ} = \frac{BC}{AC}

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

\frac{\sqrt{3}}{2} = \frac{8}{y}

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

y = \frac{8 \times 2}{\sqrt{3}}

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

y = \frac{16}{\sqrt{3}}

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

x+y = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}}

$text{x+y}=8/sqrt3+16/sqrt3$

[from (1) and (2)]

$text{ [from (1) and (2)]}$

x+y = \frac{8 + 16}{\sqrt{3}}

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

x+y = \frac{24}{\sqrt{3}}

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

x+y = \frac{24 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}

$text{x+y}=(24timessqrt3)/(sqrt3timessqrt3)$

[Rationalisation]

$text{ [Rationalisation]}$

x+y = \frac{24 \times \sqrt{3}}{3}

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

x+y = 8 \sqrt{3}

$text{x+y}=8sqrt3$

Answer:

$text{Answer:}$

The height of the tree is 8 \sqrt{3} m

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

Q3. A contractor plans to install two slides for the children to play in a park. For the children below the age of

5

$5$ years, she prefers to have a slide whose top is at a height of

1.5 m

$1.5 m$ , and is inclined at an angle of

30^{\circ}

$30^circ$ to the ground, whereas for elder children, she wants to have a steep slide at a height of

3 m

$3m$ , and inclined at an angle of

60^{\circ}

$60^circ$ to the ground. What should be the length of the slide in each case?

Sol.

$text{Sol.}$

first case

$text{first case}$

Let the length of the slide x

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

In right △ ABC

$text{In right } triangletext{ABC}$

sin 30^{\circ} = \frac{BC}{AC}

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

\frac{1}{2} = \frac{1.5}{x}

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

x = 1.5 \times 2

$text{x}=1.5times2$

$text{x}=3text{ m}$

$text{Second Case}$

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

$text{In right }triangletext{PQR}$

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

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

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

$text{y}=(3times2timessqrt3)/(sqrt3timessqrt3)$

$text{ [Rationalisation]}$

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

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

$text{Answer:}$

$text{The length of slide in first case 3 m }$

$text{and second case 2}sqrt3 text{ m}$ .

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)

$text{h}timestext{h}=9text{ tan }theta times4text{ cot }theta$

$text{ [from (1) and (2)]}$

$text{h}^2=36$

$text{h}=sqrt(36)$

$text{h}=6text{ m}$

$text{Answer:}$

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

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10th Maths 9.1

Chapter 9

Some Applications of Trigonometry

Subscribe to Quiz10 for Important MCQs

NCERT Class 10th Maths Projects

NCERT Class 10th solution of Exercise 8.1

NCERT Class 10th solution of Exercise 10.1

8th, 9th,

Exercise 9.1

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Exercise 9.1

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Some Applications of Trigonometry

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