Class 10 Maths Chapter 8 Introduction to Trigonometry Objective Questions

 

Introduction

1. 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, ∠A + ∠B + ∠C =180 °

∠A + ∠C= 180° – 90 ° = 90° ⇒ None of the angles can be ≥ 90 °

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

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

Answer: (D) 5 cm

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

Hypotenuse2 = 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 = 

$\begin{array}{l}\sqrt{3^2-4^2}\end{array}$

= 5 cm.

3. 3. 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= Sum of squares of other 2 sides

That is, 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

172≠22+92

92≠52+42

272≠132+152

So, A is the correct option.

4. 4. 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,

82≠42+62

172≠22+92

262=242+102

302≠272+132

Therefore, 24, 10 and 26 are Pythagorean triplets.

Trigonometric Identities

5. 5. If secθ + tanθ = x, then tanθ is:
      1. (x2-1) / 2x
      2. (x2+1) / 2x
      3. (x2-1) / x
      4. (x2+1) / x

Answer: (A) (x2-1) / 2x

Solution: We know that sec2θ – tan2θ = 1

Therefore, (secθ + tanθ) (secθ – tanθ) = 1

Since, (secθ + tanθ) = x

Thus, (secθ – tanθ) = 1/x

Solving both equations

We get tan θ = (x2-1) / 2x.

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

Answer: (C) p/q

∴ sin θ = p/q

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

Answer: (A) 23/8

= 15/17

sec A = 17/15

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

= 1 + (15/8)

= 23/8

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

8. 8. (sin A−2 sin3A)/ (2 cos3A−cos A)=
      1. tan A
      2. cot A
      3. sec A
      4. 1

Answer: (A) tan A

Solutions: (sin A−2 sin3A)/ (2 cos3A−cos A) = (sin A (1−2 sin2A))/ (cos A(2 cos2A−1))

= (sin A (sin2A+cos2A−2 sin2A)) / (cos A (2 cos2A− (sin2A+cos2A))

= (sin A (cos2A−sin2A)) / (cos A (cos2A−sin2A))

= tan A

Trigonometric Ratios

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

Answer: (B) 2 sin A

Solution: (cos A/cot A) + sin A

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

= sin A + sin A

= 2 sin A

10. 10. If 5tanθ=4, then value of (5sinθ -4cosθ)/(5sinθ +4cosθ) is:
        1. 1/6
        2. 5/6
        3. 0
        4. 5/3

Answer: (C) 0

Solution: Divide both numerator and denominator by cos θ and solve

(5 sin θ- 4 cos θ)/ (5 sin θ + 4 cos θ)

 

11. 11. In △PQR, PQ = 12 cm, PR = 13 cm and ∠Q=90°. Find tan P – cot R.
        1. –(119/60)
        2. 119/60
        3. 0
        4. 1

Answer: (C) 0

Solution:

Given that in △ PQR, PQ = 12 cm and PR = 13 cm.

Now, from the Pythagoras theorem,

PQ2+QR2=PR2

⇒QR2=PR2−PQ2

⇒QR2=132−122

⇒QR2=169−144=25

⇒QR= √25 = 5 cm

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

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

∴ tan P−cot R= (5/12)-(5/12) = 0

12. 12. If tanθ= (x sinϕ) / (1−xcosϕ) and, tan ϕ = (y sin θ)/ (1−y cos θ) then x/y =
        1. sinθ / (1−cosϕ)
        2. sinθ / (1−cosθ)
        3. sinθ/sinϕ
        4. sinϕ / sinθ

Answer: (C) sinθ/sinϕ

Solution: We have, tanθ = (x sinϕ)/ (1−xcosϕ)

⇒ (1−xcos ϕ) / (x sin ϕ) = 1/ tanθ ⇒ (1/ xsin ϕ) −cotϕ=cotθ

⇒ 1/ xsin ϕ= =cot θ+cot ϕ

And tan ϕ = y sinθ / (1−y cosθ) ⇒ (1−y cosθ)/ y sinθ = 1/ tan ϕ

⇒ (1/y sin θ) – cot θ = cotϕ⇒ (1/ y sin θ) =cot ϕ+cot θ

⇒ (1/y sin θ) = (1/ x sin ϕ) ⇒ x/y = sin θ/ sin ϕ

Trigonometric Ratios of Complementary Angles

13. 13. The value of tan1° × tan2° × tan3°  ×……× tan 89° is:
        1. ½
        2. 2
        3. 1
        4. 0

Answer: (C) 1

Solution: tanθcotθ=1,

tan (90−θ) =cotθ

and tan45°=1

Given: tan1°.tan2°,tan3° …….tan88°. tan89°

= (tan1°. tan89°),(tan2°. tan88°)…..(tan44°.tan46°) (tan45°)

= [(tan1°. tan (90°−1°)]. [(tan 2°. tan(90°−2°)]………. [(tan44°. tan(90°−44°)].1

= (tan1°. cot1°). (tan2°. cot2°) ……. (tan44°. cot44°)

= 1

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

Answer: (C) 36°

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

⇒ tan 2A = tan (90 – (A – 18°)

⇒ tan 2A = tan (108° – A)

⇒ 2A = 108° – A

⇒ 3A = 108°

⇒ A = 36°

15. 15. If tan 4θ=cot(θ−10°), where 4θ and(θ−10°) are acute angles, then the value of θ in degrees is:
        1. 16°
        2. 20°
        3. 32°
        4. 40°

Answer: (B) 20°

Solution: Given, tan 4θ=cot(θ−10°)

This can be written as

cot(90°− 4θ)=cot(θ−10°) —–(i)

∵ Tan θ = Cot(90°− θ)

Hence, from (i), we have

⇒90°− 4θ= θ−10°

⇒5θ = 100°

⇒θ = 20°

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

16. 16. 1. None of these
        2. C and A
        3. A and B
        4. B and C

Answer: (B) C and A

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

∠A+∠C=90°; hence angles A and C are complementary.

Trigonometric Ratios of Specific Angles

17. 17. Which of the following is correct for some θ, such that 0° ≤θ< 90°?
        1. 1/ cos θ < 1
        2. sec θ = 0
        3. 1/ sec θ < 1
        4. 1/ sec θ> 1

Answer: (C) 1/ sec θ < 1

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

18. 18. The value cot2 30°−2cos2 60°−3/4sec2 45°−4sin2 30° is:
        1. 2
        2. -1
        3. 1
        4. 0

Answer: (D) 0

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

0°<A+B≤90°, find A and B.

19. 19. 1. 25°,35°
        2. 30°, 30°
        3. 45°, 15°
        4. 10°,50°

Answer: (C) 45°, 15°

Solution: If A+B lies in this range 0°<A+B≤90°

cosec (A+B) =

$\begin{array}{l}\frac{2}{\sqrt{3}}\end{array}$

only when A+B=60° …….. (1)

sec (A-B) =

$\begin{array}{l}\frac{2}{\sqrt{3}}\end{array}$

 only when A−B=30° ……..(2)

By solving equation 1 and equation 2,

A = 45° and B = 15°

20. 20. cos 1° × cos 2° × cos 3° ×……..× cos 180° is equal to:
        1. 0
        2. 1
        3. ½
        4. -1

Answer: (A) 0

Solution: Since cos 90° = 0

The given expression,

cos 1° × cos 2° × cos 3° ×….× cos 90° ×……..× cos 180° reduces to 0 as it contains cos 90°, which is equal to 0.