CBSE Class 11 Mathematics Chapter 3 Trigonometric Functions Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Trigonometric Functions with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Trigonometric Functions MCQs with Answers to know their preparation level.
Students who are searching for NCERT MCQ Questions for Class 11 Mathematics Trigonometric Functions with Answers are compiled here to get good practice on all fundamentals. Know your preparation level on MCQ Questions for Class 11 Mathematics with Answers. You can also verify your answers from our provided MCQ Class 11 Mathematics Trigonometric Functions with Answers. So, ace up your preparation with MCQ of Chapter 3 Mathematics Objective Questions.
MCQ Class 11 Mathematics Trigonometric Functions with Answers - Set - 1
Question 1:
If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is
(a) π/3
(b) π/2
(c) 2π/3
(d) 3π/2
Correct Answer – (C)
Given, the sides of a triangle are 13, 7, 8
Since greatest side has greatest angle,
Now Cos A = (b² + c² – a²)/2bc
⇒ Cos A = (7² + 8² – 13²)/(2×7×8)
⇒ Cos A = (49 + 64 – 169)/(2×7×8)
⇒ Cos A = (113 – 169)/(2×7×8)
⇒ Cos A = -56/(2×56)
⇒ Cos A = -1/2
⇒ Cos A = Cos 2π/3
⇒ A = 2π/3
So, the greatest angle is
= 2π/3
Question 2 :
The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is
(a) 30°
(b) 90°
(c) 60°
(d) 120°
Correct Answer – (B)
Let the lengths of the sides if ∆ABC be a, b and c
Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3
⇒ (sinA + sinB + sinC) = ( a + b + c)/2
⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2
From sin formula,Using
sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2
Now, sinB/b = 1/2
Given b = 2
So, sinB/2 = 1/2
⇒ sinB = 1
⇒ B = π/2
Question 3 :
If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is
(a) 4 : (√5 – 1)
(b) 5 : 4
(c) (√5 – 1) : 4
(d) none of these
Correct Answer – (A)
Given, the angles of a triangle be in the ratio 1 : 4 : 5
⇒ x + 4x + 5x = 180
⇒ 10x = 180
⇒ x = 180/10
⇒ x = 18
So, the angle are: 18, 72, 90
Since a : b : c = sin A : sin B : sin C
⇒ a : b : c = sin 18 : sin 72 : sin 90
⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1
⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4
Now, c /a = 4/(√5 – 1)
⇒ c : a = 4 : (√5 – 1)
Question 4 :
The value of cos 5π is
(a) 0
(b) 1
(c) -1
(d) None of these
Correct Answer – (C)
Hint:Given, cos 5π = cos (π + 4π) = cos π = -1
Question 5 :
If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is
(a) a² + b² + c²
(b) a² – b² – c²
(c) a² – b² + c²
(d) a² + b² – c²
Correct Answer – (D)
We have
(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²
MCQ Class 11 Mathematics Trigonometric Functions with Answers
Question 6 :
If 3 × tan(x – 15) = tan(x + 15), then the value of x is
(a) 30
(b) 45
(c) 60
(d) 90
Correct Answer – (B)
Given, 3×tan(x – 15) = tan(x + 15)
⇒ tan(x + 15)/tan(x – 15) = 3/1
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1)
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2
⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2
⇒ sin 2x/sin 30 = 2
⇒ sin 2x/(1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45
Question 7 :
The value of cos 180° is
(a) 0
(b) 1
(c) -1
(d) infinite
Correct Answer – (C)
180 is a standard degree generally we all know their values but if we want to go theoretically then
cos(90 + x) = – sin(x)
So, cos 180 = cos(90 + 90)
= -sin 90
= -1 {sin 90 = 1}
So, cos 180 = -1
Question 8 :
In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals
(a) none of these
(b) c/a
(c) 1
(d) a/c
Correct Answer – (C)
Given cosec A (sin B cos C + cos B sin C)
= cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1
Question 9 :
If cos a + 2cos b + cos c = 2 then a, b, c are in
(a) 2b = a + c
(b) b² = a × c
(c) a = b = c
(d) None of these
Correct Answer – (A)
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b
Question 10 :
The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is
(a) sin (x + y)
(b) sin² (x + y)
(c) sin³ (x + y)
(d) sin4 (x + y)
Correct Answer – (B)
Hint:
cos² x + cos² y – 2cos x × cos y × cos(x + y)
{since cos(x + y) = cos x × cos y – sin x × sin y }
= cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y)
= cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y
= cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y
= (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y
= cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y
= sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 )
= sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y
= (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y
= (sin x × cos y + sin y × cos x)²
= {sin (x + y)}²
= sin² (x + y)
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2.1
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2.2
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2.3
- NCERT Solutions Class 11 Mathematics Relations and Functions : Exercise 2 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
