Sine Rule MCQ Quiz - Objective Question with Answer for Sine Rule - Download Free PDF

Last updated on Jun 14, 2025

Latest Sine Rule MCQ Objective Questions

Sine Rule Question 1:

Comprehension:

Consider the following for the two (02) items that follow:
In a triangle ABC, two sides BC and CA are in the ratio 2:1 and their opposite corresponding angles are in the ratio 3: 1.

Consider the following statements:

I. The triangle is right-angled.

II. One of the sides of the triangle is 3 times the other.

III. The angles A, C and B of the triangle are in AP.

Which of the statements given above is/are correct?

  1. I only
  2. II and III only
  3. I and III only
  4. I, II and III

Answer (Detailed Solution Below)

Option 3 : I and III only

Sine Rule Question 1 Detailed Solution

Explanation:

We are given the triangle with angles:

B=x=30

A=3x=90

C=180120=60

Step 1: Check if the sum of the angles is 180°:

A+B+C=90+30+60=180

This confirms that the angles satisfy the angle sum property of a triangle.

Statement I. The triangle is right-angled.

SinceA=90, the triangle is right-angled.

Statement III: III. The angles A, C and B of the triangle are in AP.

The angles 3060,and90 are in Arithmetic Progression because:

6030=30and9060=30

This confirms that the angles are in AP.

Statement II is not correct because there is no mention of a side being 3 times the other.

∴ The correct answer is Option (I) and (III) are correct.

Hence, the correct answer is Option 3.

Sine Rule Question 2:

Comprehension:

Consider the following for the two (02) items that follow:
In a triangle ABC, two sides BC and CA are in the ratio 2:1 and their opposite corresponding angles are in the ratio 3: 1.

One of the angles of the triangle is

  1. 45
  2. 75

Answer (Detailed Solution Below)

Option 2 :

Sine Rule Question 2 Detailed Solution

Calculation:

qImage6847cc1c6f6487de7b2f5188

 

We are given the equation for the ratio of sides using the Sine Rule:

asin(3x)=bsin(x)

 

ab=sin(3x)sin(x)

Step 3: Use the identity for sin(3x), which is sin(3x)=3sin(x)4sin3(x), and substitute it into the equation:

2=3sin(x)4sin3(x)sin(x)

2sin(x)=3sin(x)4sin3(x)

sin(x)+4sin3(x)=0

sin(x)(4sin2(x)1)=0

We have two possible solutions for this equation:

sin(x)=0, which gives x = 0 (not valid in this case).

4sin2(x)1=0, which simplifies to:

sin2(x)=14sin(x)=12

x=30

∴ The correct answer is Option (2)

Sine Rule Question 3:

The sides of a triangle are in A.P, and the greatest angle is double of smallest angle. Then the sum of ratio of its sides is 

Answer (Detailed Solution Below) 15

Sine Rule Question 3 Detailed Solution

Calculation

Let the sides be ad,a,a+d.

It is understood that a>d>0 and from the figure, C is greatest and A is smallest.

By given condition, C=2A.

And hence B=π(A+C)=π3A.

Hence by sine rule we have,

a+dsinC=adsinA=asinB

or a+dsin2A=adsinA=asin(π3A).

2cosA=a+dad and aad=sin3AsinA.

aad=34sin2A=34+(2cosA)2.

aad=1+(a+dad)2=4ad(ad)2.

a0 ad=4d or a=5d.

sides are ad,a,a+d

or 4d,5d,6d.

Hence the required ratio is 4:5:6.

Then the sum of ratio of its sides is  = 15


qImage679d25d3e44deca37f008d3c

Sine Rule Question 4:

In ΔABC; with usual notations, if cosA=sinBsinC, then the triangle is _______.

  1. Acute angled triangle
  2. Equilateral triangle
  3. Obtuse angled triangle
  4. Right angled triangle

Answer (Detailed Solution Below)

Option 4 : Right angled triangle

Sine Rule Question 4 Detailed Solution

We know that 2RsinB=b=AC and 2RsinC=c=AB.

Thus, cosA=ACAB.

Using above relation we can claim that ABC is a right angled triangle with AB as hypotenuse, right angled at C.

Sine Rule Question 5:

In ABC if sin2A+sin2B=sin2C and l(AB)=10, then the maximum value of the area of ABC is

  1. 50
  2. 102
  3. 25
  4. 252

Answer (Detailed Solution Below)

Option 3 : 25

Sine Rule Question 5 Detailed Solution

Calculation

sin2A+sin2B=sin2C

a2+b2=c2 (Sine Rule)

A(ABC)=12ab.....(1)

From sine rule asinA=bsinB=csinC

asinA=bsinB=101

a=10sinA,b=10sinB

Using equation (1)

A(ABC)=12(10sinA)(10sinB)

=50sinAsinB

But maximum value of sinAsinB=12

Maximum value of A(ABC)=50×12=25

OR

C=90ABC is right angled triangle

Area of is maximum when it is 454590.

A(ABC)=12×52×52=25


qImage671b410bd66cd79e6a2c06df

Hence option 3 is correct

Top Sine Rule MCQ Objective Questions

If sin (C + D) = √3/2 and sec (C - D) = 2/3 then what is the value of C and D?

  1. 45 degree and 15 degree
  2. 30 degree and 30 degree
  3. 15 degree and 30 degree
  4. 60 degree and 30 degree

Answer (Detailed Solution Below)

Option 1 : 45 degree and 15 degree

Sine Rule Question 6 Detailed Solution

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Given:

 sin (C + D) = √3/2

sec (C - D) = 2/√3

Calculations:

If  sin (C + D) = √3/2 and sec (C - D) = 2/√3

Then,

⇒ C + D = 60°.............(1)

C - D = 30°..............(2)

Solving 1 & 2 .

C = 45°

D = 15°

∴ Option 1 is the correct answer.

If the angle of triangle A, B and C are in AP and b : a = √3 : 1, then what is the value of A?

  1. 30° 
  2. 45° 
  3. 60° 
  4.  90° 

Answer (Detailed Solution Below)

Option 1 : 30° 

Sine Rule Question 7 Detailed Solution

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Concepts:

Sine law ⇔ asinA=bsinB=csinC

Where a, b and c are sides and A, B and C are angles.

F1 A.K 6.5.20 Pallavi D 3

Here; Side a faces angle A, side b faces angle B and side c faces angle C

Calculation:

Here, A, B and C are in AP

So, 2B = A + C

And sum of angles of triangle, A + B + C = 180°

⇒ 2B + B = 180 

⇒ 3B = 180 

⇒ B = 60° 

Now, by sine rule,

asinA=bsinB

ba=sinBsinA

31=sin60osinA

⇒ sin A = (√3/2) × (1/√3)

⇒ sin A = 1/2

⇒ A = 30° 

Hence, option (1) is correct. 

In a triangle ABC if a = 2, b = 4 and sin A = 1/4, then what is angle B equal to?

  1. π6
  2. π3
  3. π4
  4. π2

Answer (Detailed Solution Below)

Option 1 : π6

Sine Rule Question 8 Detailed Solution

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Concept:

Sine Rule:

sinAa=sinBb=sinCc

Where a, b and c are sides and A, B and C are angles.

F1 Pranali Defence 23.06.22 D1 V1

Here; Side a faces angle A, side b faces angle B and side c faces angle C

Calculation:

Given: a = 2, b = 4 and sin A = 1/4

sinAa=sinBb

⇒ 1/42=sinB4

⇒ sin B = 12

⇒ B = 30∘ π6

Additional Information

Cosine Rule:

cos A = b2+c2a22bc

cos B = c2+a2b22ac

cos C = a2+b2c22ab

In ΔABC if sin2 A + sin2 B = sin2 C and l(AB)=10  then the maximum value of the area of ΔABC is 

  1. 50
  2. 102
  3. 25
  4. 252

Answer (Detailed Solution Below)

Option 3 : 25

Sine Rule Question 9 Detailed Solution

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Concept:

In ΔABC, sine rule

asinA=bsinB=csinC

A(ΔABC)=12×base×height

 

Calculations:

Given, In ΔABC if sin2 A + sin2 B = sin2 C and

length of Side (AB) = 10 ⇒ c = 10

F5 Madhuri SSC 26.08.2022 D1

⇒ a+ b= c2

Pythagoreans theorem is verified.

C=90

Hence,  ΔABC is right angled triangle.

A(ΔABC)=12×base×height

A(ΔABC)=12×a×b....(1)

By sine rule, we have

asinA=bsinB=csinC

asinA=bsinB=10sin90

asinA=bsinB=101

a=10sinA and b=10sinB

 Equation (1) becomes, 

A(ΔABC)=12×10sinA×10sinB

A(ΔABC)=50×sinA×sinB ....(2)

Since, C=90and sum of all angles of triangle is 180º

(A+B)=90

⇒Max (sinA×sinB) = 12

Equation (2) becomes,

A(ΔABC)=50×12

A(ΔABC)=25

Consider the following statements:

1. If ABC is a right-angled triangle, right-angled at A, and if sin B=13, then cosec C = 3.

2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.

Which of the above statements is/are correct?

  1. 1 only
  2. 2 only
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 2 : 2 only

Sine Rule Question 10 Detailed Solution

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Concept:

Pythagoras theorem: In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Sine Rule in triangle ABC, having sides a, b, c:

F1 Amar Madhuri 02.02.2022 D1

sinAa=sinBb=sinCc=1K

sin A - cos B = 2cos(A+B2)sin(AB2)

Calculation:

1. We have,

sin B=13=PH

Hwence, Right angle triangle can be drawn as,

F1 Aman Shraddha 11.09.2020 D2

Here, P = 1
H = 3
B = x

Using Pythagoras theorem,

32 = 12 + x2

x2 = 8

x=22

Now , Cosec C = 322

Hence Ststement (1) is not correct.

2. Given,

b cos B = c cos C

Using the sine rule,

sinAa=sinBb=sinCc=1K

⇒ b = K sin B & c = K sin C

⇒ 2 sin B cos B = 2 sin C cos C

Assume,
K = 2

⇒ 2 sin B cos B = 2 sin C cos C

⇒ sin 2B = sin 2C  [2 sin x cos x = sin 2x]

⇒ sin 2B - sin 2C = 0

Using Formula:

sin C - sin D = 2 cos (C+D2) × sin (CD2)

⇒ 2 cos (B + C) sin (B - C) = 0

In this case,

Either,

cos (B + C) = 0

Hence, (B + C) = 90° .... (1)

Or,

Or sin (B -C) = 0

B - C = 0

B = C  .... (2)

From equation (1) & (2),

B = C = 45°

So ABC must be issosceles 

Hence, option (2) is correct.

In a triangle ABC, side c = 2, angle A = 45°, side a=22, then what is angle C equal to?

  1. 30° 
  2. 15° 
  3. 45° 
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : 30° 

Sine Rule Question 11 Detailed Solution

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Concept:

In triangle ABC, According to Sine rule

asinA=bsinB=csinC Where, a, b, c are sides and A, B, C are angles of triangle.

 

Calculation:

In a triangle ABC, c = 2, A = 45°, a=22,

Using sine rule in given triangle,

 asinA=csinC22sin45=2sinCsinC=sin452=122sinC=12C=30

 

Hence, option (1) is correct.

If any ΔABC , ∠C = 75°, ∠B = 45° and a = √3, then find the value of b.

  1. 1
  2. √2
  3. 2√2
  4. 2√3

Answer (Detailed Solution Below)

Option 2 : √2

Sine Rule Question 12 Detailed Solution

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Concept:

Concept:

In triangle ABC, According to the Sine rule

asin A=bsin B=csin C

F1 Madhuri Defence 18.07.2022 D2

Where, a, b, c are sides and A, B, C are angles of triangle.

Sum of the angle in triangle is 180° 

Calculation:

Given: ∠C = 75°, ∠B = 45° and a = √3

∠A + ∠B + ∠C = 180° 

⇒ ∠A = 180° - 75° - 45° 

⇒ ∠A = 60° 

Sine rule

asinA=bsinB

⇒ 3sin 600 = bsin 450

⇒ 332 = b12

⇒ b =22

⇒ b = √2 

In a triangle ABC, a = (1 + √3) cm, b = 2 cm and angle C = 60°, then the other two angles are

  1. 45° and 75°
  2. 30° and 90°
  3. 105° and 15°
  4. 100° and 20°

Answer (Detailed Solution Below)

Option 1 : 45° and 75°

Sine Rule Question 13 Detailed Solution

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Concept:

Consider a triangle ABC,

F2 A.K 31.7.20 Pallavi D1

Cosine rule:

cosC=a2+b2c22ab

Sine rule:

asinA=bsinB=csinC

Calculation:

Given: a = (1 + √ 3) cm, b = 2 cm and ∠C = 60°

We know that, cosC=a2+b2c22ab

cos60=(1+3)2+22c22(1+3)(2)

⇒ 2(1 + √3) = 1 + 3 + 2√3 + 4 – c2

⇒ c2 = 6

⇒ c = √6

Using sin rule we get,

asinA=bsinB=csinC

1+3sinA=2sinB=6sin60

⇒ sin B = (2 / √6) × sin 60°

sinB=12

⇒ B = 45°

Now, sum of all angles of triangles = 180°

⇒ A + B + C = 180°

⇒ A = 180° – 45° – 60°

⇒ A = 75°

Hence, other angles are 45° and 75°.

In a triangle ABC if a = 3, b = 4 and sin A = 3/4, then what is angle B equal to?

  1. π/4
  2. π/2
  3. π/3
  4. π/6

Answer (Detailed Solution Below)

Option 2 : π/2

Sine Rule Question 14 Detailed Solution

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Concept:

Sine Rule:

sinAa=sinBb=sinCc

Where a, b and c are sides and A, B and C are angles.

F1 A.K 6.5.20 Pallavi D 3

Here; Side a faces angle A, side b faces angle B and side c faces angle C

Calculation:

Given:

a = 3, b = 4 and sin A = 3/4

sinAa=sinBb

3/43=sinB4

sin B = 1

B = 90∘ = π/2

 

Additional Information

Cosine Rule:

cos A = b2+c2a22bc

cos B = c2+a2b22ac

cos C = a2+b2c22ab

Consider the following statements:

1. There exists no triangle ABC for which sin A + sin B = sin C.

2. If the angles of a triangle are in the ratio 1 : 2 : 3, then its sides will be in the ratio 1 : √3 : 2.

Which of the above statements is/are correct?

  1. 1 only
  2. 2 only
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 3 : Both 1 and 2

Sine Rule Question 15 Detailed Solution

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Concept:

a sinA=bsinB=csinC=k

 

Calculations:

Consider, a, b, c are the sides of triangle and A, B, C are the angles of the triangle. 

We know that, a sinA=bsinB=csinC=k

 sinA=ak, sinB=bk, sinC=ck

Given , sin A + sin B = sin C

⇒ A + B = C.

⇒ a + b = c

This is not posible.  

Hence, there exists no triangle ABC for which sin A + sin B = sin C.

(2) Given, the angles of a triangle are in the ratio 1 : 2 : 3.

Consider the angles of a triangle are A = x, B = 2x and C = 3x.

We know that, Sum of the angles of the triangle is 180°

Hence, x + 2x + 3x = 180

6x = 180°

x = 30º

Angles are A = 30º, B = 60º and C =  90º

We know that, a sinA=bsinB=csinC=k

  a sin30=bsin60=csin90=k

  a12=b32=c1=k

  a1=b3=c2=k 

Hence, the angles of a triangle are in the ratio 1 : 2 : 3, then its sides will be in the ratio 1 : √3 : 2..

 

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