MCQs On Introduction to trigonometry: Introduction & Explanation

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Jasmine Grover

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Trigonometry is all about the study of triangles. Pythagoras was a Greek philosopher who lived before 2500 years ago. He has contributed to many mathematical discoveries, one of which is known as the Pythagoras theorem. It is the building block of modern trigonometry. This theorem applies only to right-angled triangles (triangles having one angle 90 degrees), it says that the square of the hypotenuse of the triangle is equal to the sum of squares of the other two sides of the triangle.

Ques: What is the value of tan 60°/cot 30°

  1. 3
  2. 1
  3. 2
  4. 0

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Ans. b) 1

Explanation: The value of tan 60° = √3 and cot 30° = √3

Dividing , tan 60°/ cot 30° = √3/√3

= 1 (Ans)

Ques: (Sin 30° + cos 60°) - (sin 60° + cos 30°) equals

  1. 1 + √3
  2. 0
  3. 1 - √3
  4. 1 + 2√3

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Ans. c) 1 - √3

Explanation: The value of sin 30° = ½ , sin 60° =√3/2, cos 30° = √3/2, cos 60° = ½

Putting these values in the equation,

( ½ + ½ ) - ( √3/2 + √3/2 )

= 1 - (2 * √3/2)

1 - √3 (Ans)

Ques: What is the value of 1 - cos2x ?

  1. Sin2x
  2. Tan2x
  3. 1 - sin2x
  4. sec2x

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Ans. a) Sin2x

Explanation: From trignometric identities,

We know that cos2x + sin2x = 1

We get, 1 - cos2x = sin2x

Ques: If cosX = 2/3 . Find the value of tanX

  1. √5 /2
  2. 5/2
  3. 5 /√2
  4. √(5/2)

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Ans a) √5/2

Explanation: From trignometric identities we know,

1 + tan2x = sec2x

We also know that, cosx = 1 /secx

We get, 1 + tan2x = 1/cos2x

=> tan2x = (1/cos2x) - 1

=> tanx = √( (1/cos2x) -1 )

=> tanx = √ ( 9/4 - 1)

=> tanx = √(5/4)

=> tanx = √5 /4

Ques: If cosx = a/b , find the value of sinx

  1. √(b-a)/a
  2. (b2-a2)/b
  3. √(b2-a2)/b
  4. (b-a)/a

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Ans. c) √(b2-a2)/b

Explanation: given cosx = a/b

From trigonometric identities we know that,

sin2x + cos2x = 1

sin2x = 1 - cos2x

sinx = √(1 - cos2x)

sinx = √(1 - a2/b2)

sinx = √(b2 - a2)/b

Ques: 2 tan 30°/(1 + tan230°) is equal to

  1. sin 30°
  2. sin 60°
  3. cos 30°
  4. cos 60°

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Ans b) sin 60°

Explanation: We know the value, tan 30° = 1/√3

Putting this in the equation we get,

2 * 1/√3 / ( 1 + (1/√3) 2 )

= 2/√3 / (1 + 1/3)

= 2/√3 /((3+1)/3)

= 2/√3 / (4/3)

= 2/√3 * ¾

= √3 / 2

= sin 60°

Ques: Find the value of (sin 45° + cos 45°)

  1. 1/√2
  2. ½
  3. √3/2
  4. √2

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Ans d) √2

Explanation: we know the value of sin45° and cos45°,

sin 45° + cos 45°

= 1/√2 + 1/√2

= ( 1 + 1 )/√2

= 2 / √2

= √2

Ques: If the value of sinA = ½ . Find the value of cotA

  1. 1/√3
  2. √3
  3. 1
  4. √3/2

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Ans b) √3

Explanation: Given sin A = ½

By trigonometric identities,

cos2A + sin2A = 1

cos2A = 1 - sin2A

cosA = √(1 - sin2A)

cosA = √(1 - (1/2)2)

cosA = √(1 - (¼))

cosA = √((4-1) / 4)

cosA = √(3/4)

cosA = √3/2

Now , cotA = cosA / sinA

cotA = (√3/2) / (½)

cotA = √3

Ques: if a triangle ABC is right-angled at C. What will be the value of cos(A+B)

  1. 1
  2. 0
  3. √3/2
  4. 1/2

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Ans b) 0

Explanation: Given to us is a right-angled triangle having angle C = 90°

We know that in a triangle, all the three angles sum upto 180°.

∠A + ∠B + ∠C = 180°

∠A + ∠B = 180° - ∠C

∠A + ∠B = 180° - 90°

We get ∠A + ∠B = 90°

Now taking cos both sides

cos(A+B) = cos 90°

cos(A+B) = 0

Ques: Find the value of (tan1° tan2° tan3°… tan89°) is

  1. ½
  2. 1
  3. 0
  4. 2

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Ans b) 1

Explanation:

tan1° tan2° tan3°… tan89°

= (tan1° tan2° tan3°..tan44°). (tan45). (tan46° tan 47°… tan89°)

= (tan1° tan2° tan3°..tan44°). (tan45).(tan(90° - 44°) tan(90°- 43°).....tan(90°-1°))

= [(tan1° * cot1°).(tan2° * cot2°)......(tan44° * cot(44°)] . 1

=1 * 1* 1 * 1 *1 * 1* 1 * 1* 1 ……* 1 (because cotx = 1/tanx )

=1

Ques: Find the value of expression [cosec(75° + θ) - sec(15° - θ) - tan(55° + θ) + cot(35° - θ)]

  1. 0
  2. 1
  3. -1
  4. 3/2

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Ans a) 0

Explanation: [cosec(75° + θ) - sec(15° - θ) - tan(55° + θ) + cot(35° - θ)]

= [cosec(90° - (15° - θ)) - sec( 15° - θ) - tan(55° - θ) + cot(90° - (55 °+ θ))]

We know that cosec (90° - x ) = secx and cot(90° - x) = tanx

= [sec(15° - θ) - sec (15° - θ) - tan(55° + θ) + tan(55° + θ)]

= 0

Ques: If cos(α + β) = 0, then sin(α – β) can be reduced to

  1. Cos β
  2. Cos 2β
  3. Sin α
  4. Sin 2α

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Ans b) cos 2α

Explanation: Given that cos(α + β) = 0

We get cos(α + β) = cos90 (cos 90 = 0)

α + β = 90

α = 90 - β

Now , sin(α - β) = sin(90 - β - β) (α = 90 - β)

= sin(90 - 2β)

= cos 2β

Ques: if sinA + sin2A = 1, then the value of expression (cos2A + cos4A)

  1. 1
  2. 2
  3. 3
  4. ½

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Ans. a) 1

Explanation: Given,

sinA + sin2A = 1

sinA = 1 - sin2A

sinA = cos2A (sin2A + cos2A = 1)

Squaring both sides

sin2A = cos4A

1 - cos2A = cos4A ( sin2A + cos2A = 1)

1 = cos2A + cos4A

cos2A + cos4A = 1

Ques: If cos 9α = sinα and 9α < 90°. Find the value of tan 5α

  1. 0
  2. √3
  3. 1/√3
  4. 1

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Ans d) 1

Explanation: Given,

cos 9α = sinα and 9α < 90°

It implies that 9α is acute angle

Cos 9α = cos (90° - α) (cos(90 - α) = sinα)

We get

9α = 90° - α

10α = 90°

α = 9°

Tan 5α = tan 5 * 9 = tan45° = 1

Ques: Find the value of the expression (sin6θ + cos6θ + 3 sin2θ cos2θ )

  1. 0
  2. 1
  3. 2
  4. 3

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Ans b) 1

Explanation: we know that sin2x + cos2x = 1

Cubing this equation,

(Sin2x + cos2x)3 = 1

(sin2x )3 + (cos2x)3 + 3sin2x cos2x (sin2x + cos2x ) = 1

Sin6x + cos6x + 3sin2x cos2x = 1

Ques: sin2A = 2sinA only when the value of A is

  1. 0
  2. 45
  3. 60
  4. 30

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Ans a) 0

Explanation: given

sin2A = 2 sinA

When A = 0

sin2A = sin 0 = 0

And, 2sinA = 2* sin0 = 0

LHS = RHS

Ques: In a triangle ABC, right angled at B, AB = 24cm, BC = 7cm. Find the value of tanC

  1. 12/7
  2. 24/7
  3. 7/24
  4. 20/7

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Ans b) 24/7

Explanation: given AB = 24cm and BC = 7cm

The given triangle is right-angled at B

We get tanC = opposite side / adjacent side

tanC = 24 /7

Related links:

CBSE X Related Questions

1.

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

      2.

      Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

          3.

          Solve the following pair of linear equations by the substitution method. 
          (i) x + y = 14 
              x – y = 4   

          (ii) s – t = 3 
              \(\frac{s}{3} + \frac{t}{2}\) =6 

          (iii) 3x – y = 3 
                9x – 3y = 9

          (iv) 0.2x + 0.3y = 1.3 
               0.4x + 0.5y = 2.3 

          (v)\(\sqrt2x\) + \(\sqrt3y\)=0
              \(\sqrt3x\) - \(\sqrt8y\) = 0

          (vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
              \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

              4.
              Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) \(x + y = 5\),\( 2x + 2y = 10\) (ii)\( x – y = 8 , 3x – 3y = 16\) (iii) \(2x + y – 6 = 0\) , \(4x – 2y – 4 = 0\) (iv) \(2x – 2y – 2 = 0,\) \( 4x – 4y – 5 = 0\)

                  5.
                  Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
                  (i) 2, 4, 8, 16, . . . .
                  (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
                  (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
                  (iv) – 10, – 6, – 2, 2, . . .
                  (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
                  (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
                  (vii) 0, – 4, – 8, –12, . . . .
                  (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
                  (ix) 1, 3, 9, 27, . . . .
                  (x) a, 2a, 3a, 4a, . . . .
                  (xi) a, \(a^2, a^3, a^4,\)  . . . .
                  (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
                  (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
                  (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
                  (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

                      6.
                      If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos2 A – sinA or not

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