Concise Selina Solutions for Class 9 Maths Chapter 24 - Solution Of Right Triangles

Selina Solutions for Class 9 Maths Chapter 24, Solution Of Right Triangles, are provided here. It is important for students to understand the concepts taught in Class 9 as they are continued in class 10 syllabus as well. To score good marks in Class 9 Maths examination, it is advised to solve the entire questions provided in each exercise across all the chapters in the book by Selina publication. The Selina solutions for Class 9 Maths helps students in understanding all the concepts in a better way. Download pdf of Class 9 Maths Chapter 24 Selina Solutions from the given link.

Download PDF of Selina Solutions for Class 9 Maths Chapter 24:-Download Here

Exercise 24

1. Find ‘x’ if:

Solution:

(i) From the figure, we have

sin 60^o = 20/x

√3/2 = 20/x

∴ x = 40/√3

(ii) From the figure, we have

tan 30^o = 20/x

1/√3 = 20/x

∴ x = 20√3

(iii) From the figure, we have

sin 45^o = 20/x

1/√2 = 20/x

∴ x = 20√2

2. Find angle ‘A’ if:

Solution:

(i) From the figure, we have

cos A = 10/20

= ½

cos A = cos 60^o

Hence,

A = 60^o

(ii) From the figure, we have

sin A = (10/√2)/10

= 1/√2

sin A = sin 45^o

Hence,

A = 45^o

(iii) From the figure, we have

tan A = (10√3)/10

= √3

tan A = tan 60^o

Hence,

A = 60^o

3. Find angle ‘x’ if:

Solution:

The given figure is drawn as follows:

We have,

tan 60^o = 30/AD

√3 = 30/AD

AD = 30/√3

Again,

sin x = AD/20

AD = 20 sin x

Now,

20 sin x = 30/√3

sin x = 30/20√3

sin x = 3/2√3

sin x = √3/2

sin x = sin 60^o

⇒ x = 60^o

4. Find AD, if:

Solution:

(i) In right ∆ABE, we have

tan 45^o = AE/BE

1 = AE/BE

AE = BE

Thus, AE = BE = 50 m

Now,

In rectangle BCDE, we have

DE = BC = 10 m

Thus, the length of AD is given by

AD = AE + DE

= 50 + 10

= 60 m

(ii) In right ∆ABD, we have

sin B = AD/AB

sin 30^o = AD/100 [As ∠ACD is the exterior angle of ∆ABC]

½ = AD/100

⇒ AD = 50 m

5. Find the length of AD.

Given: ∠ABC = 60^o, ∠DBC = 45^o and BC = 40 cm

Solution:

In right ∆ABC, we have

tan 60^o = AC/BC

√3 = AC/40

AC = 40√3 cm

Next,

In right ∆BDC, we have

tan 45^o = DC/BC

1 = DC/40

DC = 40 cm

Now, from the figure it’s clearly seen that

AD = AC – DC

= 40√3 – 40

= 40(√3 – 1)

Hence, the length of AD is 29.28 cm

6. Find the lengths of diagonals AC and BD. Given AB = 60 cm and ∠BAD = 60^o.

Solution:

We know that, diagonals of a rhombus bisect each other at right angles and also bisect the angle of vertex.

Considering the figure as shown below:

Now, we have

OA = OC = ½ AC,

OB = OD = ½ BD

∠AOB = 90^o and

∠OAB = 60^o/2 = 30^o

Also given that AB = 60 cm

Now,

In right ∆AOB, we have

sin 30^o = OB/AB

½ = OB/60

OB = 30

Also,

cos 30^o = OA/AB

√3/2 = OA/60

OA = 51.96 cm

Therefore,

Length of diagonal AC = 2 x OA

= 2 x 51.96

= 103.92 cm

Length of diagonal BD = 2 x OB

= 2 x 30

= 60 cm

7. Find AB.

Solution:

Considering the given figure, let’s construct FP ⊥ ED

Now,

In right ∆ACF, we have

tan 45^o = 20/AC

1 = 20/AC

AC = 20

Next,

In right ∆DEB, we have

tan 60^o = 30/BD

√3 = 30/BD

BD = 30/√3

= 17.32 cm

Also, given FC = 20 and ED = 30

So, EP = 10 cm

Thus,

tan 60^o = FP/EP

√3 = FP/10

FP = 10√3

= 17.32 cm

And, FP = CD

Therefore, AB = AC + CD + BD

= 20 + 17.32 + 17.32

= 54.64 cm

8. In trapezium ABCD, as shown, AB || DC, AD = DC = BC = 20 cm and ∠A = 60^o. Find:

(i) length of AB

(ii) distance between AB and DC

Solution:

Constructing two perpendiculars to AB from the point D and C respectively.

Now, since AB||CD we have PMCD as a rectangle

Considering the figure,

(i) From right ∆ADP, we have

cos 60^o = AP/AD

½ = AP/20

AP = 10

Similarly,

In right ∆BMC, we have

BM = 10 cm

Now, from the rectangle PMCD we have

CD = PM = 20 cm

Therefore,

AB = AP + PM + MB

= 10 + 20 + 10

= 40 cm

(ii) Again, from the right ∆APD, we have

sin 60^o = PD/20

√3/2 = PD/20

PD = 10√3

Hence, the distance between AB and CD is 10√3 cm

9. Use the information given to find the length of AB.

Solution:

In right ∆AQP, we have

tan 30^o = AQ/AP

1/√3 = 10/AP

AP = 10√3

Also,

In right ∆PBR, we have

tan 45^o = PB/BR

1 = PB/8

PB = 8

Therefore, AB = AP + PB

= 10√3 + 8

10. Find the length of AB.

Solution:

In right ∆ADE, we have

tan 45^o = AE/DE

1 = AE/30

AE = 30 cm

Also, in right ∆DBE, we have

tan 60^o = BE/DE

√3 = BE/30

BE = 30√3 cm

Therefore, AB = AE + BE

= 30 + 30√3

= 30(1 + √3) cm

11. In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB.

Given that ∠AED = 60^o and ∠ACD = 45^o. Calculate:

(i) AB (ii) AC (iii) AE

Solution:

(i) In right ∆ADC, we have

tan 45^o = AD/DC

1 = 2/DC

DC = 2 cm

And, as AD || DC and AD ⊥ EC, ABCD is a parallelogram

So, opposite sides are equal

Hence, AB = DC = 2 cm

(ii) Again, in right ∆ADC

sin 45^o = AD/AC

1/√2 = AD/AC

AC = 2√2 cm

(iii) In right ∆ADE, we have

sin 60^o = AD/AE

√3/2 = 2/AE

AE = 4/√3

12. In the given figure, ∠B = 60^°, AB = 16 cm and BC = 23 cm,

Calculate:

(i) BE (ii) AC

Solution:

In ∆ABE, we have

sin 60^o = AE/AB

√3/2 = AE/16

AE = √3/2 x 16

= 8√3 cm

(i) In ∆ABE, we have ∠AEB = 90^°

So, by Pythagoras Theorem, we get

BE² = AB² – AE²

BE² = (16)² – (8√3)²

BE² = 256 – 192

BE² = 64

Taking square root on both sides, we get

BE = 8 cm

(ii) EC = BC – BE

= 23 – 8

= 15

In ∆AEC, we have

∠AEC = 90^° 

So, by Pythagoras Theorem, we get

AC² = AE² + EC²

AC² = (8√3)² + (15)²

AC² = 192 + 225

AC² = 147

Taking square root on both sides, we get

AC = 20.42 cm

13. Find

(i) BC

(ii) AD

(iii) AC

Solution:

(i) In right ∆AEC, we have

tan 30^o = AB/BC

1/√3 = 12/BC

BC = 12√3 cm

(ii) In right ∆ABD, we have

cos A = AD/AB

cos 60^o = AD/AB

½ = AD/12

AD = 12/2

AD = 6 cm

(iii) In right ∆ABC, we have

sin B = AB/AC

sin 30^o = AB/AC

½ = 12/AC

AC = 12 x 2

AC = 24 cm

14. In right-angled triangle ABC; ∠B = 90^o. Find the magnitude of angle A, if:

(i) AB is √3 times of BC

(ii) BC is √3 times of AB

Solution:

Considering the figure below:

(i) We have, AB is √3 times of BC

AB/BC = √3

cot A = cot 30^o

A = 30^o

Hence, the magnitude of angle A is 30^o

(ii) Again, from the figure

BC/AB = √3

tan A = √3

tan A = tan 60^o

A = 60^o

Hence, the magnitude of angle A is 60^o

15. A ladder is placed against a vertical tower. If the ladder makes an angle of 30^o with the ground and reaches upto a height of 15 m of the tower; find length of the ladder.

Solution:

Given that the ladder makes an angle of 30^o with the ground and reaches upto a height of 15 m of the tower

Let’s consider the figure shown below:

Let’s assume the length of the ladder is x metre

Now, from the figure we have

15/x = sin 30^o [As perpendicular/hypotenuse = sine]

15/x = ½

x = 30 m

Hence, the length of the ladder is 30 m.

16. A kite is attached to a 100 m long string. Find the greatest height reached by the kite when its string makes an angles of 60^o with the level ground.

Solution:

Given that the kite is attached to a 100 m long string and it makes an angle of 60^o with the ground level

Let’s consider the figure shown below:

Let’s assume the greatest height to be x metre

Now, from the figure we have

x/100 = sin 60^o [As perpendicular/hypotenuse = sine]

x/100 = √3/2

x = 100 x (√3/2)

= 86.6 m

Hence, the greatest height reached by the kite is 86.6 m.

17. Find AB and BC, if:

Solution:

(i) Let assume BC to be x cm

BD = BC + CD = (x + 20) cm

Now,

In ∆ABD, we have

tan 30^o = AB/BD

1/√3 = AB/(x + 20)

x + 20 = √3AB        … (1)

And,

In ∆ABC, we have

tan 45^o = AB/BC

1 = AB/x

AB = x       … (2)

Now, using (1) in (2) we get

AB + 20 = √3AB

AB(√3 – 1) = 20

AB = 20/(√3 – 1) 

= 20/(√3 – 1) x [(√3 + 1)/( √3 + 1)]

= 20(√3 + 1)/ (3 – 1)

     = 27.32 cm

Hence from (2), we have

AB = BC = x = 27.32 cm

Therefore, AB = 27.32cm, BC = 27.32cm

(ii) Let’s assume BC to be x m

BD = BC + CD

= (x + 20) cm

In ∆ABD, we have

tan 30^o = AB/BD

1/√3 = AB/(x + 20)

x + 20 = √3 AB … (1)

In ∆ABC, we have

tan 60^o = AB/BC

√3 = AB/x

x = AB/√3 … (2)

Now, from (1) we have

AB/√3 + 20 = √3AB

AB + 20√3 = 3AB

2AB = 20√3

AB = 20√3/2

= 10√3

= 17.32 cm

And, from (2) we get

x = AB/√3

= 17.32/√3

= 10 cm

Hence, BC = x = 10cm

Therefore,

AB = 17.32 cm and BC = 10cm

(iii) Let’s assume BC to be x cm

BD = BC + CD

= (x + 20) cm

In ∆ABD, we have

tan 45^o = AB/BD

1 = AB/(x + 20)

x + 20 = AB … (1)

Also,

In ∆ABD, we have

tan 60^o = AB/BC

√3 = AB/x

x = AB/√3 … (2)

Now, from (1) we have

AB/√3 + 20 = AB

AB + 20√3 = √3AB

AB (√3 – 1) = 20√3

AB = 20√3/(√3 – 1)

On rationalizing, we get

AB = 20√3(√3 + 1)/(3 -1)

= 10√3(√3 + 1)

= 47.32 cm

And, from (2) we get

x = AB/√3

= 47.32/√3

= 27.32 cm

Hence, BC = x = 27.32 cm

Therefore,

AB = 47.32 cm and BC = 27.32 cm

18. Find PQ, if AB = 150 m, ∠P = 30^o and ∠Q = 45^o.

Solution:

(i) In ∆APB, we have

tan 30^o = AB/PB

1/√3 = 150/PB

PB = 150√3

= 259.80 m

Also, in ∆ABQ

tan 45^o = AB/BQ

1 = 150/BQ

BQ = 150 m

Therefore,

PQ = PB + BQ

= 259.80 + 150

= 409.80 m

(ii) In ∆APB, we have

tan 30^o = AB/PB

1/√3 = 150/PB

PB = 150√3

= 259.80 m

Also, in in ∆APB, we have

tan 45^o = AB/BQ

1 = 150/BQ

BQ = 150 cm

Therefore, PQ = PB – BQ

= 259.80 – 150

= 109.80 m

19. If tan x^o = 5/12, tan y^o = ¾ and AB = 48 m; find the length of CD.

Solution:

Given,

tan x^o = 5/12, tan y^o = ¾ and AB = 48 m;

Let’s consider the length of BC to x metre

In ∆ADC, we have

tan x^o = DC/AC

5/12 = DC/(48 + x)

5(48 + x) = 12DC

240 + 5x = 12DC … (1)

Also, in ∆BDC we have

tan y^o = CD/BC

¾ = CD/x

x = 4CD/3 … (2)

Also, from (1) we get

240 + 5(4CD/3) = 12CD

240 + 20CD/3 = 12CD

720 + 20CD = 36CD

36CD – 20CD = 720

16CD = 720

CD = 720/16

CD = 45

Therefore, the length of CD is 45 m.

20. The perimeter of a rhombus is 96 cm and obtuse angle of it is 120^o. Find the lengths of its diagonals.

Solution:

As rhombus has all sides equal

Let’s consider the diagram as below:

Hence, PQ = 96/4 = 24 cm

And, ∠PQR = 120^o

We also know that,

In a rhombus, the diagonals bisect each other perpendicularly and the diagonal bisect the angle at vertex

Thus, ∠POR is a right-angle triangle

∠PQR = ½ (∠PQR)

= 60^o

sin 60^o = perpendicular/hypotenuse

√3/2 = PO/PQ

√3/2 = PO/24

PO = 24√3/2

= 12√3

= 20/784 cm

Hence,

PR = 2PO

= 2 x 20.784

= 41.568 cm

Also, we have

cos 60^o = base/hypotenuse

½ = OQ/24

OQ = 24/2

= 12 cm

Hence, SQ = 2 x OQ

= 2 x 12

= 24 cm

Therefore, the length of the diagonals PR is 41.568 cm and of SQ is 24 cm

Selina Solutions for Class 9 Maths Chapter 24-Solution Of Right Triangles

The Chapter 24, Solution Of Right Triangles, contains 1 exercise and the Selina Solutions given here has the answers to all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

24.1 Introduction

24.2 Simple 2-D problems involving one right angled triangle.

Selina Solutions for Class 9 Maths Chapter 24-Solution Of Right Triangles

To solve a right angled triangle means, to find the values of remaining angles and remaining sides under two conditions, namely, when one side and one acute angle are given or when two sides of the triangle are given. Chapter 24 of class 9 gives the students an overview of the different problems related to the Solution Of Right Triangles. Read and learn Chapter 24 of Selina textbook to learn more about Solution Of Right Triangles along with the concepts covered in it. Solve the Selina Solutions for Class 9 effectively to score high in the examination.