# 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.

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Exercise 24

1. Find ‘x’ if:

Solution:

(i) From the figure, we have

sin 60o = 20/x

âˆš3/2 = 20/x

âˆ´ x = 40/âˆš3

(ii) From the figure, we have

tan 30o = 20/x

1/âˆš3 = 20/x

âˆ´ x = 20âˆš3

(iii) From the figure, we have

sin 45o = 20/x

1/âˆš2 = 20/x

âˆ´ x = 20âˆš2

2. Find angle ‘A’ if:

Solution:

(i) From the figure, we have

cos A = 10/20

= Â½

cos A = cos 60o

Hence,

A = 60o

(ii) From the figure, we have

sin A = (10/âˆš2)/10

= 1/âˆš2

sin A = sin 45o

Hence,

A = 45o

(iii) From the figure, we have

tan A = (10âˆš3)/10

= âˆš3

tan A = tan 60o

Hence,

A = 60o

3. Find angle ‘x’ if:

Solution:

The given figure is drawn as follows:

We have,

Again,

Now,

20 sin x = 30/âˆš3

sin x = 30/20âˆš3

sin x = 3/2âˆš3

sin x = âˆš3/2

sin x = sin 60o

â‡’ x = 60o

Solution:

(i) In right âˆ†ABE, we have

tan 45o = 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

= 50 + 10

= 60 m

(ii) In right âˆ†ABD, we have

sin 30o = AD/100 [As âˆ ACD is the exterior angle of âˆ†ABC]

5. Find the length of AD.

Given:Â âˆ ABC = 60o, âˆ DBC = 45o and BC = 40 cm

Solution:

In right âˆ†ABC, we have

tan 60o = AC/BC

âˆš3 = AC/40

AC = 40âˆš3 cm

Next,

In right âˆ†BDC, we have

tan 45o = DC/BC

1 = DC/40

DC = 40 cm

Now, from the figure itâ€™s clearly seen that

= 40âˆš3 – 40

= 40(âˆš3 – 1)

Hence, the length of AD is 29.28 cm

Â

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

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 = Â½ AC,

OB = OD = Â½ BD

âˆ AOB = 90o and

âˆ OAB = 60o/2 = 30o

Also given that AB = 60 cm

Now,

In right âˆ†AOB, we have

sin 30o = OB/AB

Â½ = OB/60

OB = 30

Also,

cos 30o = OA/AB

âˆš3/2 = 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, letâ€™s construct FP âŠ¥ ED

Now,

In right âˆ†ACF, we have

tan 45o = 20/AC

1 = 20/AC

AC = 20

Next,

In right âˆ†DEB, we have

tan 60o = 30/BD

âˆš3 = 30/BD

BD = 30/âˆš3

= 17.32 cm

Also, given FC = 20 and ED = 30

So, EP = 10 cm

Thus,

tan 60o = FP/EP

âˆš3 = FP/10

FP = 10âˆš3

= 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 âˆ A = 60o. 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 âˆ†ADP, we have

Â½ = AP/20

AP = 10

Similarly,

In right âˆ†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 âˆ†APD, we have

sin 60o = PD/20

âˆš3/2 = PD/20

PD = 10âˆš3

Hence, the distance between AB and CD isÂ 10âˆš3 cm

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

Solution:

In right âˆ†AQP, we have

tan 30o = AQ/AP

1/âˆš3 = 10/AP

AP = 10âˆš3

Also,

In right âˆ†PBR, we have

tan 45o = PB/BR

1 = PB/8

PB = 8

Therefore, AB = AP + PB

= 10âˆš3 + 8

10. Find the length of AB.

Solution:

tan 45o = AE/DE

1 = AE/30

AE = 30 cm

Also, in right âˆ†DBE, we have

tan 60o = BE/DE

âˆš3 = BE/30

BE = 30âˆš3 cm

Therefore, AB = AE + BE

= 30 + 30âˆš3

= 30(1 + âˆš3) 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 thatÂ âˆ AED = 60oÂ andÂ âˆ ACD = 45o. Calculate:

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

Solution:

(i) In right âˆ†ADC, we have

1 = 2/DC

DC = 2 cm

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

So, opposite sides are equal

Hence, AB = DC = 2 cm

AC = 2âˆš2 cm

(iii) In right âˆ†ADE, we have

âˆš3/2 = 2/AE

AE = 4/âˆš3

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

Calculate:

(i) BEÂ (ii) AC

Solution:

In âˆ†ABE, we have

sin 60o = AE/AB

âˆš3/2 = AE/16

AE = âˆš3/2 x 16

= 8âˆš3 cm

(i) In âˆ†ABE,Â we have âˆ AEBÂ = 90Â°

So, by Pythagoras Theorem, we get

BE2Â = AB2Â – AE2

BE2Â = (16)2Â – (8âˆš3)2

BE2Â = 256 – 192

BE2Â = 64

Taking square root on both sides, we get

BE = 8 cm

(ii) EC = BC – BE

= 23 – 8

= 15

In âˆ†AEC,Â we have

âˆ AECÂ = 90Â°Â

So, by Pythagoras Theorem, we get

AC2Â = AE2Â + EC2

AC2Â = (8âˆš3)2Â + (15)2

AC2Â = 192 + 225

AC2Â = 147

Taking square root on both sides, we get

AC = 20.42 cm

13. Find

(i) BC

(iii) AC

Solution:

(i) In right âˆ†AEC,Â we have

tan 30o = AB/BC

1/âˆš3 = 12/BC

BC = 12âˆš3 cm

(ii) In right âˆ†ABD,Â we have

(iii) In right âˆ†ABC,Â we have

sin B = AB/AC

sin 30o = AB/AC

Â½ = 12/AC

AC = 12 x 2

AC = 24 cm

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

(i) AB isÂ âˆš3 times of BC

(ii) BC isÂ âˆš3 times of AB

Solution:

Considering the figure below:

(i) We have, AB isÂ âˆš3 times of BC

AB/BC = âˆš3

cot A = cot 30o

A = 30o

Hence, the magnitude of angle A is 30o

(ii) Again, from the figure

BC/AB = âˆš3

tan A = âˆš3

tan A = tan 60o

A = 60o

Hence, the magnitude of angle A is 60o

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

Letâ€™s consider the figure shown below:

Letâ€™s assume the length of the ladder is x metre

Now, from the figure we have

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

15/x = Â½

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 60oÂ with the level ground.

Solution:

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

Letâ€™s consider the figure shown below:

Letâ€™s assume the greatest height to beÂ xÂ metre

Now, from the figure we have

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

x/100 = âˆš3/2

x = 100 x (âˆš3/2)

= 86.6 m

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

17. Find AB and BC, if:

Solution:

(i) Let assumeÂ BC to be x cm

BD = BC + CD =Â (x + 20) cm

Now,

InÂ âˆ†ABD, we have

tan 30o =Â AB/BD

1/âˆš3 = AB/(x + 20)

x + 20 =Â âˆš3AB Â  Â  Â Â Â … (1)

And,

InÂ âˆ†ABC, we have

tan 45o =Â AB/BC

1 = AB/x

AB = x Â  Â  Â Â … (2)

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

ABÂ + 20 =Â âˆš3AB

AB(âˆš3 – 1) = 20

ABÂ = 20/(âˆš3 – 1)Â

= 20/(âˆš3 – 1) x [(âˆš3 + 1)/( âˆš3 + 1)]

= 20(âˆš3 + 1)/ (3 â€“ 1)

Â  Â  Â = 27.32Â cm

Hence from (2), we have

AB = BC = x = 27.32 cm

Therefore,Â ABÂ = 27.32cm,Â BCÂ = 27.32cm

(ii) Letâ€™s assume BC to be x m

BD = BC + CD

= (x + 20) cm

In âˆ†ABD, we have

tan 30oÂ = AB/BD

1/âˆš3 = AB/(x + 20)

x + 20 = âˆš3 AB â€¦ (1)

Â

InÂ âˆ†ABC, we have

tan 60o =Â AB/BC

âˆš3 = AB/x

x = AB/âˆš3 â€¦ (2)

Now, from (1) we have

AB/âˆš3 + 20 = âˆš3AB

AB + 20âˆš3 = 3AB

2AB = 20âˆš3

AB = 20âˆš3/2

= 10âˆš3

= 17.32 cm

And, from (2) we get

x = AB/âˆš3

= 17.32/âˆš3

= 10 cm

Hence,Â BCÂ =Â xÂ = 10cm

Therefore,

ABÂ = 17.32 cm and BCÂ = 10cm

Â

(iii) Letâ€™s assume BC to be x cm

BD = BC + CD

=Â (x + 20) cm

InÂ âˆ†ABD, we have

tan 45o = AB/BD

1 = AB/(x + 20)

x + 20 = AB â€¦ (1)

Also,

InÂ âˆ†ABD, we have

tan 60o = AB/BC

âˆš3 = AB/x

x = AB/âˆš3 â€¦ (2)

Â

Now, from (1) we have

AB/âˆš3 + 20 = AB

AB + 20âˆš3 = âˆš3AB

AB (âˆš3 – 1) = 20âˆš3

AB = 20âˆš3/(âˆš3 – 1)

On rationalizing, we get

AB = 20âˆš3(âˆš3 + 1)/(3 -1)

= 10âˆš3(âˆš3 + 1)

= 47.32 cm

And, from (2) we get

x = AB/âˆš3

= 47.32/âˆš3

= 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,Â âˆ P = 30oÂ andÂ âˆ Q = 45o.

Solution:

(i) In âˆ†APB, we have

tan 30o = AB/PB

1/âˆš3 = 150/PB

PB = 150âˆš3

= 259.80 m

Also, in âˆ†ABQ

tan 45o = AB/BQ

1 = 150/BQ

BQ = 150 m

Therefore,

PQ = PB + BQ

= 259.80 + 150

= 409.80 m

(ii) In âˆ†APB, we have

tan 30o = AB/PB

1/âˆš3 = 150/PB

PB = 150âˆš3

= 259.80 m

Also, inÂ in âˆ†APB, we have

tan 45o = AB/BQ

1 = 150/BQ

BQ = 150 cm

Therefore, PQ = PB â€“ BQ

= 259.80 â€“ 150

= 109.80 m

19. If tan xoÂ = 5/12, tan yoÂ =Â Â¾ and AB = 48 m; find the length of CD.

Solution:

Given,

tan xoÂ = 5/12, tan yoÂ = Â¾ and AB = 48 m;

Letâ€™s consider the length of BC to x metre

tan xo = DC/AC

5/12 = DC/(48 + x)

5(48 + x) = 12DC

240 + 5x = 12DC â€¦ (1)

Also, inÂ âˆ†BDC we have

tan yo = CD/BC

Â¾ = CD/x

x = 4CD/3 â€¦ (2)

Also, from (1) we get

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

240 + 20CD/3 = 12CD

720 + 20CD = 36CD

36CD â€“ 20CD = 720

16CD = 720

CD = 720/16

CD = 45

Therefore, the length of CD is 45Â m.

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

Solution:

As rhombus has all sides equal

Letâ€™s consider the diagram as below:

Hence, PQ = 96/4 = 24 cm

And, âˆ PQR = 120o

We also know that,

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

Thus, âˆ POR is a right-angle triangle

âˆ PQR = Â½ (âˆ PQR)

= 60o

sin 60o = perpendicular/hypotenuse

âˆš3/2 = PO/PQ

âˆš3/2 = PO/24

PO = 24âˆš3/2

= 12âˆš3

= 20/784 cm

Hence,

PR = 2PO

= 2 x 20.784

= 41.568 cm

Also, we have

cos 60o = base/hypotenuse

Â½ = OQ/24

OQ = 24/2

= 12 cm

Hence,Â 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.