NCERT Solutions For Class 10 Maths Chapter 9

NCERT Solutions Class 10 Maths Some Applications of Trigonometry

Maths requires skills and understanding, along with lot of practice. It is essential for the students to have a grip in mathematics, which can help in their day to day life. To have a better understanding of the topic, and ease to students, BYJU’S provide students of class 10 with NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry. Students can refer to these solution, which is provided by our experts. Here students can find easy methods of solving a particular question that gives them a good exposure and knowledge.Students can also download the NCERT Solutions for class 10 maths chapter 9 pdf which can be saved for later use.

NCERT solutions for class 10 maths chapter 9 pdf are prepared by our subject experts under the guidelines of NCERT to assists students in their board exam preparations.

NCERT Solutions Class 10 Maths Chapter 9 Exercises

1. If sun’s elevation is 45°. Find the length of the shadow of a pole of height 8 m.

In rt. Δ PQR, B=90° and QPR=45°

1

∴ tan 45° = QRPQ

1 = 8PQ

AB=81 = 8m

 

2. A pole 8 metre high cast a shadow 8 m on the ground. Find the sun’s elevation.

In rt. Δ PQR, B=90°

2

∴ PQ = 8 m

QR = 8 m

Let QPR=θ

∴ tan θ = QRPQ

= QRPQ = 1

θ = 45°

 

3. If the angle of elevation of the top of a hill from two points at distances p and q from the base and in the same straight line with it are complementary. Find the height of the hill.

Let   PQ = h be the height of the hill,

QR = b, QS = a, such that PSQ=θ

PRQ=90°θ

3

In rt. Δ PQS, we have

ha=tan θ

h = a tan θ          …..(1)

In rt. Δ PQR, we have

hb=tan(90°θ) = cot θ

h = b cot θ         …..(2)

Multiplying (1) and (2), we get

h2 = ab tan θ cot θ = ab

h=ab

 

4. From the figure, find the angle of depression of point Z from the point R.

4

YRS=XYR (alternative.int. a )

YRS = 60°

ZRS+YRZ=60° ZRS+30°=60° ZRS=30°

 

5. Find the angle of elevation of the moon when the shadow of a tree h metres high is 3h metres long.

5

The height of the pole be h m and length of its shadow is 3h

Let θ be the elevation of the sun.

Consider rt.agl Δ ABC

tan θ = h3h=13

= tan 30°

   θ = 30°

 

 6. A ladder 14 metres long just reaches the top of a wall. If the ladder makes an angle of 45° with the wall, find the height of the wall.

Here, length of the ladder is 14m and angle of elevation is 90°-45°=45°.

Let h m be the height of the wall

6

Consider rt.agl Δ PQR

QRPR = sin 30°

h14=22

h = 1422 m

∴ The length of the wall is 72

 

 7. An observer 1.8 m tall is 20.8 m away from a tower 22.6 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

7

Here, PQ be the observer and RS be the tower of height 24 m.

PQ = TS = 1.8 m

RT = RS – TS

= 22.6 – 1.8

= 20.8 m

Now, Consider rt.agl ΔPTR, we have

RTAT = tan θ

20.820.8 = tan θ

1 = tan θ

tan 45° = tan θ

θ =45°

 

8.From a balloon vertically above a straight road, the angles of depression of two cars at an instant is found to be 45° and 60°. If the car are 100 m apart, find the height of the balloon.

Let the height of the balloon at P be h metres. Let A and B be the two cars. Thus, AB = 100m.

8

∠ PAQ= 45° and ∠ PBQ=60°

Now, Consider rt.agl ΔAQP,

PQAQ = tan 45°

         PQAQ = 1

         PQ = AQ= h m

Now, Consider rt.agl ΔPBQ,

PQBQ = tan 60°

PQBQ = 3

hh100 = 3

h = 3h3(100)

(31)h = 3(100)

h = 3(100)31

h = 3(100)31 ×3+13+1

h = 3(100)(3+1)2

= 50(3+3)

 

 9. The shadow of a tower on a level surface is 40 m long when moon’s elevation is 30° then when it is 60°.Find the height of the tower.

XY=40 m, RXS =30°, RYS = 60°

9

Now, Consider rt.agl ΔRYS

RYSY = tan 60°

         RYSY = 3

                                   h = 3 SY

Consider rt.agl ΔRXQ

RXSX = tan 30°

    RXXY+YS = 13

    h40+YS     = 13

    3h = 40 + YS

    3h = 40 + h3

3h = 403 + h

2h = 403

h = 203

∴ The height of the tower is 203 m

 

10. From the top of a tree h m high, the angle of depression of two objects, which are in the line with foot of the tree are γ and δ (δ> γ). Find the distance between the two objects.

Let RS be the tree of height h m, X and Y are two objects such that

RXS = γ , RYS = δ

10

Now, Consider rt.agl ΔYSR

RSYS = tan δ

    RS = YS tan δ     …..(i)

Now, Consider rt.agl ΔXSR

RSXS = tan γ

RSXY+YS = tan γ

         RS = (XY+YS) tan γ     …..(ii)

htanγ = XY + RStanδ     [using (i)]

h cot γ = XY + h cot δ

XY = h (cot γ + cot δ)

 

11. The angle of elevation of the top of a tower from two distinct points x and y from its foot are complementary. Prove that the height of the tower is xy.

Let h m be the height of the tower PQ. X and Y are two distinct points from its foot, such that PX = x and PY = y

11

Now, Consider rt.agl ΔXQP, we obtain

PQPX = tan θ

hx = tan θ

h = x tan θ     …..(i)

Consider rt.agl ΔYQP, we obtain

PQPY = tan (90°θ)

hy = cot θ

h = y ×1tanθ   …..(ii)

Multiply (I) and (ii), we get

h2 = x tan θ ×y×1tanθ = xy

h = xy

∴ The height of the tower is xy.

 

12. The angle of elevation of the top of a tower from certain point at 30°. If the observer moves 10 m towards the tower, the angle of elevation of the top increases by 15° . Find the height of the tower.

Let h m be the height of the tower RS and P, Q are two points 10 m apaet, such that

SPR=30°andSQR=45°.

12

Now, Consider rt.agl ΔQRS

SRQR = tan 45°

                  SR = QR = h m

Now, Consider rt.agl ΔPRS

SRPR = tan 30°

h10+h = 13

        3 h = 10 + h

(31) h = 10

h = 1031×3+13+1 = 10(3+1)2

∴ The height of the tower is 5(3+1)

 

13. The angle of elevation of the top of a vertical tower from a point on the groung is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Let XY be the vertical tower of height h m.

C and D are two points 10 m apart such that,

13

CD = ZY = 10 m XCZ=45°andXBY=60°

XZ = XY – ZY = (h – 10) m

Consider rt.agl ΔDYX

XYDY = tan 60°

hDY = 3

DY = h3

Consider rt.agl ΔCZX

XZCZ = tan 45°

XZ = CZ

H – 10 = h3

h (113) = 10

h(313) = 10

h = 10331×3+13+1

= 10331×3+13+1

= 5(3 + 3)m

 

14. The angle of elevation of the top of a tower 27 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and the height of the other tower.

Let us assume that PQ and RS be the two tower of height  h m and 27 m respectively.

RQS = 60° PSQ = 30°

14

Consider rt.agl ΔRQS

RSQS = tan 60°

27QS = 3

QS = 273     …..(i)

Consider rt.agl ΔPSQ

PQQS = tan30°

hQS = 13

   3 h = QS     …..(ii)

From (i) and (ii), we have

3 h = 273

h = 273 = 9 m

From (ii), we get QS = 93 m

∴ The distance between two towers is 93 m and the height of  the other tower is 9 m.

 

15. The lower window of a house is at a height of 3 m above the ground and its upper window is 6 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon from the ground.

Let P and Q be the positions of two windows of a house, such that PR = 3 m, QR = 9 m. Let U be the position of balloon of height h m above the ground. UPS = 60°, UQT = 30°, US = h – 3 and UT = h – 9

15

Consider rt.agl ΔQTU

UTQT = tan 30º

h – 9 = QT ×13

3 (h – 8) = QT      …..(i)

Consider rt.agl ΔUPS

USPS = tan 60º

h – 3 = PS ×3

  h33 = PS      …..(ii)

Also, QT = PS     …..(iii)

From (i), (ii) and (iii), we get

3 (h – 9) = h33

  3h -27 = h – 3

   3h – h = 27 – 3

   2h = 24

    h = 12

∴ height of the balloon above the ground is 12 m.