# CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry Objective Questions

CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry Objective Questions teach students about the concept of trigonometry. This concept involves the study of some important ratios of the right angle triangle with respect to the angles called trigonometric ratios of the angles. From this chapter, students will learn to establish some of the trigonometric identities and solve certain specific angles (from 0Â° to 90Â°) of the trigonometric ratios, sine, cosine, tangents, and more. They can also learn other concepts like specific angles, ratios of complementary angles, and so on. In this article, they can access the solutions of all the CBSE Class 10 Maths Chapter 8 – Introduction to Trigonometry Objective Questions.

### List of Sub-Topics Covered in Chapter 8

This article comprises only the Multiple Choice Questions (MCQs) taken from various important topics in the chapter, with detailed solutions and answers. The CBSE Class 10 Maths Objective Questions in the PDF are also provided as per the topics given. As the changed exam pattern is expected to include more MCQs, these CBSE Class 10 Maths Chapter 8 – Introduction to Trigonometry Objective Questions will help the students to prepare better for the exams.

8.1 Introduction (4 MCQs from the Topic)

8.2 Trigonometric Identities (4 MCQs from the Topic)

8.3 Trigonometric Ratios (4 MCQs from the Topic)

8.4 Trigonometric Ratios of Complementary Angles (4 MCQs from the Topic)

8.5 Trigonometric Ratios of Specific Angles (4 MCQs from the Topic)

## Download CBSE Class 10 Maths Chapter 8 – Introduction to Trigonometry Objective Questions Free PDF

### Introduction

1. In a right triangle ABC, the right angle is at B. Which of the following is true about the other two angles, A and C?
1. There is no restriction on the measure of the angles
2. Both angles should be obtuse
3. Both angles should be acute
4. One of the angles is acute, and the other is obtuse

Answer: (C) Both angles should be acute

Solution: In triangle ABC,Â âˆ A +Â âˆ B +Â âˆ C =180 Â°

âˆ A +Â âˆ C= 180Â°Â – 90Â Â°Â = 90Â° â‡’Â None of the angles can beÂ â‰¥Â 90Â Â°

1. In a right triangle ABC, the right angle is at B. What is the length of the missing side in the figure?

1. 25 cm
2. 12 cm
3. 7 cm
4. 5 cm

Solution: Pythagoras theorem: In a right-angled triangle,

Hypotenuse2Â = Sum of squares of other 2 sides

That is,

c2=a2+b2

Here a = 4 cm and b = 3 cm,

So the missing side = c =Â

$$\begin{array}{l}\sqrt{3^{2}-4^{2}}\end{array}$$
= 5 cm.

1. Which of the following numbers can form sides of a right-angled triangle?
1. 13 cm, 27 cm, 15 cm
2. 4 cm, 5 cm, 9 cm
3. 2 cm, 17 cm, 9 cm
4. 10 cm, 6 cm, 8 cm

Answer: (D) 10 cm, 6 cm, 8 cm

Solution: The basic condition for any type of triangle is:

(i) The sum of 2 sides of a triangle should be greater than the third side

(ii) The difference between any 2 sides should be less than the third side

For a triangle to be a right-angled triangle, there is an additional condition.

According to the Pythagoras theorem, in a right-angled triangle,

Hypotenuse2=Â SumÂ ofÂ squaresÂ ofÂ otherÂ 2Â sides

That is,Â c2=a2+b2; also note that the hypotenuse is the largest side in a right triangle.

Considering each of the given options,

102=62+82

172â‰ 22+92

92â‰ 52+42

272â‰ 132+152

So, A is the correct option.

1. Which of the following are Pythagorean triplets?
1. 4 cm , 6 cm , 8 cm
2. 24 cm , 10 cm , 26 cm
3. 13 cm , 27 cm , 30 cm
4. 2 cm , 17 cm , 9 cm

Answer: (B) 24 cm, 10 cm, 26 cm

Solution: Pythagorean triplets are those sets of numbers that satisfy the Pythagoras theorem.

Considering the options given to us,

82â‰ 42+62

172â‰ 22+92

262=242+102

302â‰ 272+132

Therefore, 24, 10 and 26 are Pythagorean triplets.

### Trigonometric Identities

1. IfÂ secÎ¸Â +Â tanÎ¸Â = x, thenÂ tanÎ¸Â is:
1. (x2-1) / 2x
2. (x2+1) / 2x
3. (x2-1) / x
4. (x2+1) / x

Solution: We know that sec2Î¸Â –Â tan2Î¸Â = 1

Therefore, (secÎ¸Â +Â tanÎ¸) (secÎ¸Â –Â tanÎ¸) = 1

Since, (secÎ¸Â +Â tanÎ¸) = x

Thus, (secÎ¸Â –Â tanÎ¸) =Â 1/x

Solving both equations

We get tanÂ Î¸ = (x2-1) / 2x.

1. If p cotÎ¸ =
$$\begin{array}{l}\sqrt{q^{2}-p^{2}}\end{array}$$
, then the value of sinÎ¸ is ___. (Î¸ being an acute angle)
1. q/3p
2. q/2p
3. p/q
4. 0

âˆ´Â sin Î¸Â =Â p/q

1. If sin A =Â 8/17, find the value of secA cosA + cosecA cosA.
1. 23/8
2. 15/8
3. 8/15
4. 6/23

= 15/17

sec A = 17/15

secA cosA + cosecA cosA = (17/15) * (15/17) + (17/15) Â * (15/17)

= 1 + (15/8)

= 23/8

Therefore, the value of secA cosA + cosecA cosA is 23/8.

1. (sinÂ Aâˆ’2Â sin3A)/ (2Â cos3Aâˆ’cosÂ A)=
1. tan A
2. cot A
3. sec A
4. 1

Solutions: (sinÂ Aâˆ’2Â sin3A)/ (2Â cos3Aâˆ’cosÂ A) = (sinÂ A (1âˆ’2Â sin2A))/ (cosÂ A(2Â cos2Aâˆ’1))

= (sinÂ A (sin2A+cos2Aâˆ’2Â sin2A)) / (cosÂ A (2Â cos2Aâˆ’ (sin2A+cos2A))

= (sinÂ A (cos2Aâˆ’sin2A)) / (cos A (cos2Aâˆ’sin2A))

= tan A

### Trigonometric Ratios

1. (cos A/cot A) + sin A= ____________.
1. cot A
2. 2 sin A
3. 2 cos A
4. sec A

Solution: (cos A/cot A) + sin A

= Cos A / (cos A/sin A) + sin A

= sin A + sin A

= 2 sin A

1. If 5tanÎ¸=4, then value of (5sinÎ¸Â -4cosÎ¸)/(5sinÎ¸Â +4cosÎ¸) is:
1. 1/6
2. 5/6
3. 0
4. 5/3

Solution: Divide both numerator and denominator by cosÂ Î¸Â and solve

(5 sin Î¸- 4 cos Î¸)/ (5 sin Î¸ + 4 cos Î¸)

1. In â–³PQR, PQ = 12 cm, PR = 13 cm and âˆ Q=90Â°. Find tan P – cot R.
1. â€“(119/60)
2. 119/60
3. 0
4. 1

Solution:

Given that inÂ â–³Â PQR, PQ = 12 cm and PR = 13 cm.

Now, from the Pythagoras theorem,

PQ2+QR2=PR2

â‡’QR2=PR2âˆ’PQ2

â‡’QR2=132âˆ’122

â‡’QR2=169âˆ’144=25

â‡’QR= âˆš25 = 5Â cm

Now, tan P= opposite side/adjacent side = QR/PQ= 5/12

cot R= adjacent side/opposite side = QR/PQ = 5/12

âˆ´ tan Pâˆ’cot R= (5/12)-(5/12) = 0

1. If tanÎ¸= (xÂ sinÏ•) / (1âˆ’xcosÏ•) and,Â tan Ï• = (y sin Î¸)/ (1âˆ’yÂ cos Î¸) then x/y =
1. sinÎ¸ / (1âˆ’cosÏ•)
2. sinÎ¸ / (1âˆ’cosÎ¸)
3. sinÎ¸/sinÏ•
4. sinÏ• / sinÎ¸

Solution: We have, tanÎ¸ = (xÂ sinÏ•)/ (1âˆ’xcosÏ•)

â‡’ (1âˆ’xcos Ï•) / (xÂ sin Ï•) = 1/ tanÎ¸ â‡’ (1/ xsin Ï•) âˆ’cotÏ•=cotÎ¸

â‡’ 1/ xsin Ï•= =cot Î¸+cot Ï•

And tan Ï• = y sinÎ¸ / (1âˆ’y cosÎ¸) â‡’ (1âˆ’y cosÎ¸)/ y sinÎ¸ = 1/ tan Ï•

â‡’ (1/y sin Î¸) â€“ cot Î¸ = cotÏ•â‡’ (1/ y sin Î¸) =cot Ï•+cot Î¸

â‡’ (1/y sin Î¸) = (1/ x sin Ï•) â‡’ x/y = sin Î¸/ sin Ï•

### Trigonometric Ratios of Complementary Angles

1. The value of tan1Â°Â Ã—Â tan2Â°Â Ã—Â tan3Â°Â Â Ã—â€¦…Ã—Â tanÂ 89Â° is:
1. Â½
2. 2
3. 1
4. 0

Solution: tanÎ¸cotÎ¸=1,

tan (90âˆ’Î¸) =cotÎ¸

andÂ tan45Â°=1

Given:Â tan1Â°.tan2Â°,tan3Â°Â …….tan88Â°.Â tan89Â°

= (tan1Â°.Â tan89Â°),(tan2Â°.Â tan88Â°)…..(tan44Â°.tan46Â°)Â (tan45Â°)

= [(tan1Â°.Â tan (90Â°âˆ’1Â°)].Â [(tan 2Â°.Â tan(90Â°âˆ’2Â°)]………. [(tan44Â°.Â tan(90Â°âˆ’44Â°)].1

=Â (tan1Â°.Â cot1Â°).Â (tan2Â°.Â cot2Â°)Â …….Â (tan44Â°.Â cot44Â°)

= 1

1. If tan2A = cot(A-18Â°), then value of A is:
1. 27Â°
2. 24Â°
3. 36Â°
4. 18Â°

Solution: Given, tan 2A = cot (A –Â 18Â°)

â‡’Â tan 2A = tanÂ (90 – (A –Â 18Â°)

â‡’Â tan 2A = tanÂ (108Â°Â – A)

â‡’Â 2A =Â 108Â°Â – A

â‡’Â 3A =Â 108Â°

â‡’Â A =Â 36Â°

1. IfÂ tanÂ 4Î¸=cot(Î¸âˆ’10Â°), whereÂ 4Î¸Â and(Î¸âˆ’10Â°) are acute angles, then the value of Î¸ in degrees is:
1. 16Â°
2. 20Â°
3. 32Â°
4. 40Â°

Solution: Given,Â tanÂ 4Î¸=cot(Î¸âˆ’10Â°)

This can be written as

cot(90Â°âˆ’Â 4Î¸)=cot(Î¸âˆ’10Â°)Â —–(i)

âˆµ Tan Î¸ = Cot(90Â°âˆ’ Î¸)

Hence, from (i), we have

â‡’90Â°âˆ’Â 4Î¸=Â Î¸âˆ’10Â°

â‡’5Î¸ = 100Â°

â‡’Î¸ = 20Â°

1. The given triangle is right-angled at B. Which pair of angles are complementary?

1. None of these
2. C and A
3. A and B
4. B and C

Solution: Two angles are said to be complementary if their sum is 90Â°. The triangle is right-angled at B. With the angle sum property of the triangle, âˆ A+âˆ B+âˆ C=180Â°.

âˆ A+âˆ C=90Â°; henceÂ angles A and C are complementary.

### Trigonometric Ratios of Specific Angles

1. Which of the following is correct for someÂ Î¸, such thatÂ 0Â°Â â‰¤Î¸<Â 90Â°?
1. 1/ cosÂ Î¸ <Â 1
2. secÂ Î¸Â = 0
3. 1/ secÂ Î¸ <Â 1
4. 1/ secÂ Î¸> 1

Answer: (C) 1/ secÂ Î¸ <Â 1

Solution: 1/ sec Î¸ = cos Î¸. And the value of cos Î¸ ranges from 0 to 1

1. The valueÂ cot2Â 30Â°âˆ’2cos2Â 60Â°âˆ’3/4sec2Â 45Â°âˆ’4sin2Â 30Â° is:
1. 2
2. -1
3. 1
4. 0

1. If Cosec (A+ B) =
$$\begin{array}{l}\frac{2}{\sqrt{3}}\end{array}$$
sec(A-B)=
$$\begin{array}{l}\frac{2}{\sqrt{3}}\end{array}$$

0Â°<A+Bâ‰¤90Â°, find A and B.

1. 25Â°,35Â°
2. 30Â°, 30Â°
3. 45Â°, 15Â°
4. 10Â°,50Â°

Solution: If A+B lies in this rangeÂ 0Â°<A+Bâ‰¤90Â°

cosec (A+B) =

$$\begin{array}{l}\frac{2}{\sqrt{3}}\end{array}$$
only whenÂ A+B=60Â°Â …….. (1)

sec (A-B) =

$$\begin{array}{l}\frac{2}{\sqrt{3}}\end{array}$$
Â only whenÂ Aâˆ’B=30Â°Â ……..(2)

By solving equation 1 and equation 2,

A = 45Â° and B = 15Â°

1. cosÂ 1Â°Â Ã—Â cosÂ 2Â°Â Ã—Â cosÂ 3Â°Â Ã—â€¦â€¦..Ã—Â cosÂ 180Â°Â is equal to:
1. 0
2. 1
3. Â½
4. -1

Solution: SinceÂ cosÂ 90Â°Â =Â 0

The given expression,

cosÂ 1Â°Â Ã—Â cosÂ 2Â°Â Ã—Â cosÂ 3Â°Â Ã—….Ã—Â cosÂ 90Â°Â Ã—â€¦â€¦..Ã—Â cosÂ 180Â° reduces to 0 as it contains cos 90Â°, which is equal to 0.

Students can practise these multiple-choice questions in order to self-assess their knowledge of the important topics discussed in this chapter. They can download the free PDF of the objective questions from the link given above.

The solutions for these MCQs are also given in a step-by-step procedure for the students to refer to. However, students preparing for the CBSE Class 10 Maths Board examination are advised to practise the given Mathematics questions independently without referring to the answers or solutions to self-assess their performance. Here, you can also find some CBSE Class 10 Maths Chapter 8 Extra MCQs.

## CBSE Class 10 Maths Chapter 8 Extra MCQs

1.Â Â Given that cos A = 4/5, then calculate tan A.Â
(A) 3/5
(B) 3/4
(C) 4/3
(D) 4/5

2.Â (sin30Â° + cos30Â°) â€“ (sin 60Â° + cos60Â°)
(A) â€“ 1
(B) 0
(C) 1
(D) 2

All that students wish to know about the CBSE Class 10 and its study resources is available at BYJUâ€™S. Stay tuned for more updates.