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Test: Trigonometric Functions - Commerce MCQ


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25 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Trigonometric Functions

Test: Trigonometric Functions for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Trigonometric Functions questions and answers have been prepared according to the Commerce exam syllabus.The Test: Trigonometric Functions MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Trigonometric Functions below.
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Test: Trigonometric Functions - Question 1

If in a triangle ABC, (s − a) (s − b) = s (s − c), then  angle C is equal to

Detailed Solution for Test: Trigonometric Functions - Question 1

Thus, C/2 = 45º ⇒ C = 90º

Test: Trigonometric Functions - Question 2

if A = 45,B = 75, then a+c√2 is equal to,

Detailed Solution for Test: Trigonometric Functions - Question 2

∠C = 180 − 45 − 75 = 60

sin 75 = sin (45 + 30) = sin45.cos30 + sin30.cos45
sin 75 = 1/√2 x √3/2 + 1/2 1/√2


Test: Trigonometric Functions - Question 3

The angles of a triangle are as 1 : 2 : 7, then ratio of greatest side to least side is

Detailed Solution for Test: Trigonometric Functions - Question 3

Let the angles be x, 2x, and 7x respectively.
⇒ x + 2x + 7x = 180° [angle sum property of triangle]
⇒ 10x = 180°
⇒ x = 18°
Hence, the angles are: 18°,36°,126°


⇒ Greatest Side = K sin 126°
⇒ Smallest side = K sin 18°
So, required ratio =K sin 126°/ K sin 18°
= sin(90°+36°)/sin18°= cos36°/sin18°    [sin(90°+x)=cosx]
= (√5+1)/(√5−1)

Test: Trigonometric Functions - Question 4

If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is

Detailed Solution for Test: Trigonometric Functions - Question 4

a = 13, b = 7, c = 8
 By cosine formula,
⇒ A = 120º = 2π/3

*Multiple options can be correct
Test: Trigonometric Functions - Question 5

There exists a triangle ABC satisfying the conditions

Detailed Solution for Test: Trigonometric Functions - Question 5

The sine formula is 
a/sinA = b/sinB 
⇒ a sin B = b sin A
(a) b sin A = a 
⇒ a sin B = a 
⇒ B = π/2 
Since, ∠A  < π/2 therefore, the triangle is possible.
(b) b sin A < a 
⇒ a sin B < a 
⇒ sinB < 1 
⇒ ∠B exists 
Now, b > a 
⇒ B > A since A < π/2 
∴ The triangle is possible.

Test: Trigonometric Functions - Question 6

In a ΔABC, (b +c) cos A + (c + a) cos B + (a + b) cos C is equal to

Detailed Solution for Test: Trigonometric Functions - Question 6

(b+c) cos A + (c + a) cos B + (a + b) cos C
⇒ b cos A + c cos A + c cos B + a cos B + a cos C + b cos C
⇒ (b cos C + c cos B) + (c cos A + a cos C) + (a cos B + b cos A) ...(1)
Using projection formula,
a = (b cos C + c cos B) ...(2)
b = (c cos A + a cos C) ...(3)
c = (a cos B + b cos A) ...(4)

On adding the projection formula, we get the initial expression
i.e. (2) + (3) + (4) = (1)
∴ (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c

Test: Trigonometric Functions - Question 7

If the angles of a triangle ABC are in A.P., then

Detailed Solution for Test: Trigonometric Functions - Question 7

Let the angles be a,a+d,a+2d.
Then,
a+a+d+a+2d=180º
a+d=60º
So, ∠B=60º,
By cosine formula,

⇒ b= a+ c− ac

Test: Trigonometric Functions - Question 8

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

Test: Trigonometric Functions - Question 9

In a triangle ABC,  is equal to

Test: Trigonometric Functions - Question 10

The area of a triangle is 80cm2 and its perimeter is 8 cm. The radius of its inscribed circle is

Detailed Solution for Test: Trigonometric Functions - Question 10

Area of a triangle = 80 cm2, a + b + c = 8 cm
s = perimeter/2=(a + b + c)/2 = 8/2 = 4 cm

Radius of inscribed circle (r) = (area of the triange)/s = 80/4 = 20 cm

Test: Trigonometric Functions - Question 11

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

Test: Trigonometric Functions - Question 12

If a = 4, b = 3 and A = 60, then c is a root of the equation

Detailed Solution for Test: Trigonometric Functions - Question 12


c− 7 = 3c ⇒ c− 3c − 7 = 0

Replace c by x : x2 - 3x - 7 = 0

Test: Trigonometric Functions - Question 13

If b2+c= 3a2, then cot B + cot C - cot A =

Test: Trigonometric Functions - Question 14

Test: Trigonometric Functions - Question 15

If the sides of a triangle be 2x + 1, x2- 1 and x2 + x + 1, then the greatest angle is

Test: Trigonometric Functions - Question 16

If in a triangle ABC, acos2 (c/2) + ccos2 (A/2) = 3b/2, then its sides a, b, c are in

Test: Trigonometric Functions - Question 17

In a ΔABC, if a sin A = b sin B then the triangle is

Test: Trigonometric Functions - Question 18

If in a ΔABC, a cos A = b cos B, then the triangle is

Test: Trigonometric Functions - Question 19

In a ΔABC, is equal to 

Test: Trigonometric Functions - Question 20

In a triangle ABC, a = 2b and ∠A = 3∠ B then angle A is

Test: Trigonometric Functions - Question 21

In a triangle ABC, AD is the median A to BC, then its length is equal to

Test: Trigonometric Functions - Question 22

If cos A + cos B = , then the sides of the triangle ABC are in

Detailed Solution for Test: Trigonometric Functions - Question 22

cos A + cos B = 4 sin2(C/2​)
⇒ 2 cos (A+B)/2​ cos (A−B)/2 ​= 4 sin2(C/2​)

∵ A + B + C = π ⇒ A + B = π − C

⇒ cos (π−C)/2 ​cos (A−B)/2 ​= 2 sin2(C/2​)
⇒ sin C/2 ​cos (A−B)/2 ​= 2 sin2(C/2​)
⇒ cos (A−B)/2 = 2 sin (C/2​)
⇒ cos C/2 ​cos (A−B)/2 = 2 sin (C/2​) cos (C/2)​
⇒ cos (π−(A+B)​)/2 cos (A−B)/2 = sin C
⇒ 2 sin (A+B)/2 ​cos (A−B)/2​ = sin C
⇒ sin A + sin B = 2 sin C

∵ a/sinA​ = b/sinB​ = c/sinC​ = k
⇒sinA = ak, sin B = bk , sin C = ck

⇒ ak + bk = 2(ck)
⇒ a+b=2c

Therefore the sides of triangle a,b,c are in A.P.

Test: Trigonometric Functions - Question 23

ABC is an equilateral triangle of each side a (> 0). The inradius of the triangle is

Test: Trigonometric Functions - Question 24

In any ΔABC, the expression (a + b + c) (a + b – c) (b + c – a) (c + a – b) is equal to

Test: Trigonometric Functions - Question 25

In any ΔABC, if C =90, then tan B/2 is equal to

Detailed Solution for Test: Trigonometric Functions - Question 25


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