Sine Rule MCQ Quiz in मल्याळम - Objective Question with Answer for Sine Rule - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 12, 2025

നേടുക Sine Rule ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Sine Rule MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Sine Rule MCQ Objective Questions

Top Sine Rule MCQ Objective Questions

Sine Rule Question 1:

\(\frac{1 + \cos (A - B) \cos C}{1 + \cos (A - C) \cos B}\) is equal to

  1. \(\frac{a^2 + b^2}{a^2 + c^2}\)
  2. \(\frac{ b^2+ c^2}{b^2- c^2}\)
  3. \(\frac{c^2 - a^2}{a^2 + b^2}\)
  4. None of these

Answer (Detailed Solution Below)

Option 1 : \(\frac{a^2 + b^2}{a^2 + c^2}\)

Sine Rule Question 1 Detailed Solution

Concept:

Sine Rule:

In a triangle Δ ABC, 

\(\frac{sin \ A}{{a}} = \frac{sin\ B}{{b}} = \frac{sin\ C}{{c}} \)

Where, 

∠A + ∠B + ∠C = π 

Formula used:

  1. cos(π - θ) = -cos θ 
  2. 1 - sin2θ = cos2θ  
  3. cos(A + B) cos(A − B) = cos2A - sin2Bcos(A+B)cos(AB)=cos2Asin2B

 

Calculation:

We have

\(\frac{1 + \cos (A - B) \cos C}{1 + \cos (A - C) \cos B}\)

\(⇒ \frac{1 + \cos (A - B) \cos [π -(A+B)]}{1 + \cos (A - C) \cos [π-(A+C)]}\)  (∵ ∠A + ∠B + ∠C = π )

\(⇒ \frac{1 + \cos (A - B) \cos (A+B)}{1 + \cos (A -C) \cos (A+C)}\)    (∵ cos(π - θ) = -cos θ)

\(⇒ \frac{1 - [\cos^2A- \sin^2 B]}{1 - [\cos^2A- \sin^2 C]}\)

\(⇒ \frac{\sin^2A+ \sin^2 B}{sin^2A+\sin^2 C}\)       (∵1 - sin2θ = cos2θ)

From the sine rule,

\(\frac{sin \ A}{{a}} = \frac{sin\ B}{{b}} = \frac{sin\ C}{{c}} =k \)

⇒ sin A = ka, sin B = kb, sin C = kc

Hence, required value

\( \frac{\sin^2A+ \sin^2 B}{sin^2A+\sin^2 C}=\frac{k^2(a^2+b^2)}{k^2(a^2+c^2)}\)

\(\therefore\frac{1 + \cos (A - B) \cos C}{1 + \cos (A - C) \cos B}=\frac{(a^2+b^2)}{(a^2+c^2)}\)

Sine Rule Question 2:

In a triangle ABC, ∠A = 30°, ∠B = 45°, AC = 3√2 cm then find the length BC - 

  1. 3 cm
  2. 4 cm
  3. 5 cm
  4. 8 cm

Answer (Detailed Solution Below)

Option 1 : 3 cm

Sine Rule Question 2 Detailed Solution

Formula used - 

In a triangle ABC, Sine rule, 

(sinA/a) = (sinB/b) = (sinC/c)

Given - 

∠A = 30°, ∠B = 45°, AC = 3√2 cm

Solution - 

⇒ (sin45°/3√2) = (sin30°/BC)

⇒ (1/6) = {1/(2 BC)}

⇒ BC = 3cm

∴ side BC = 3 cm

Sine Rule Question 3:

If α = \(\frac{180^o}{7}\), then 3 sin α - 4 sin3 α is equal to

  1. cos 4α
  2. sin 3α
  3. cos 3α
  4. 0

Answer (Detailed Solution Below)

Option 2 : sin 3α

Sine Rule Question 3 Detailed Solution

Concept:

3 sin α - 4 sin3 α = sin 3α

Explanation:

We know,

3 sin α - 4 sin3 α = sin 3α

Since, α = \(\frac{180°}{7}\)

3 sin \(\frac{180°}{7}\) - 4 sin3 \(\frac{180°}{7}\) = sin (3× \(\frac{180°}{7}\))

⇒ 3 sin α 

Sine Rule Question 4:

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 4 Detailed Solution

Explanation:

We are given the triangle with angles:

\( \angle B = x = 30^\circ\)

\( \angle A = 3x = 90^\circ \)

\(\angle C = 180^\circ - 120^\circ = 60^\circ \)

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

\( \angle A + \angle B + \angle C = 90^\circ + 30^\circ + 60^\circ = 180^\circ \)

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

Statement I. The triangle is right-angled.

Since\( \angle A = 90^\circ \), the triangle is right-angled.

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

The angles \(30^\circ 60^\circ , \text and 90^\circ \) are in Arithmetic Progression because:

\( 60^\circ - 30^\circ = 30^\circ \quad \text{and} \quad 90^\circ - 60^\circ = 30^\circ \)

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 5:

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 5 Detailed Solution

Calculation:

qImage6847cc1c6f6487de7b2f5188

 

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

\( \frac{a}{\sin(3x)} = \frac{b}{\sin(x)} \)

 

\( \frac{a}{b} = \frac{\sin(3x)}{\sin(x)} \)

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

\( 2 = \frac{3\sin(x) - 4\sin^3(x)}{\sin(x)} \)

\( 2\sin(x) = 3\sin(x) - 4\sin^3(x) \)

\( -\sin(x) + 4\sin^3(x) = 0 \)

\( \sin(x)(4\sin^2(x) - 1) = 0 \)

We have two possible solutions for this equation:

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

\( 4\sin^2(x) - 1 = 0 \), which simplifies to:

\( \sin^2(x) = \frac{1}{4} \quad \Rightarrow \quad \sin(x) = \frac{1}{2} \)

\( x = 30^\circ \)

∴ The correct answer is Option (2)

Sine Rule Question 6:

If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of length of the sides of this triangle is:

  1. \(5 : 9 : 13\)
  2. \(5 : 6 : 7\)
  3. \(4 : 5 : 6\)
  4. \(3 : 4 : 5\)

Answer (Detailed Solution Below)

Option 3 : \(4 : 5 : 6\)

Sine Rule Question 6 Detailed Solution

\(a < b < c\) are in A.P.

\(\angle C = 2\angle A\) (Given)

\(\Rightarrow \sin C = \sin 2A\)

\(\Rightarrow \sin C = 2\sin A \cos A\)

\(\dfrac{\sin C}{\sin A} = 2\cos A\)

\(\Rightarrow \dfrac{c}{a} = 2 \dfrac{b^2 + c^2 - a^2}{2bc}\)

Put \(a = b - \lambda, c = b + \lambda, \lambda > 0\)

\(\Rightarrow \lambda = \dfrac{b}{5}\)

\(\Rightarrow a = b - \dfrac{b}{5} = \dfrac{4}{5}b, c = b + \dfrac{b}{5} = \dfrac{6b}{5}\)

\(\Rightarrow\) required ratio \(= 4 : 5 : 6\)

Sine Rule Question 7:

In \(\Delta ABC\); with usual notations, if \(\cos A=\dfrac{\sin B}{\sin C}\), 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 7 Detailed Solution

We know that \(2R\sin B = b = AC\) and \(2R\sin C = c = AB\).

Thus, \(\cos A = \dfrac{AC}{AB}\).

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

Sine Rule Question 8:

In \(\triangle ABC\) if \(\sin^{2} A+ \sin^{2}B = \sin^{2}C\) and \(l(AB) = 10\), then the maximum value of the area of \(\triangle ABC\) is

  1. \(50\)
  2. \(10\sqrt {2}\)
  3. \(25\)
  4. \(25\sqrt {2}\)

Answer (Detailed Solution Below)

Option 3 : \(25\)

Sine Rule Question 8 Detailed Solution

Calculation

\(\sin^{2} A+ \sin^{2}B = \sin^{2}C\)

\(\Rightarrow a^{2} + b^{2} = c^{2}\) (Sine Rule)

\(A(\triangle ABC) = \dfrac {1}{2} ab ..... (1)\)

From sine rule \(\dfrac {a}{\sin A} = \dfrac {b}{\sin B} = \dfrac {c}{\sin C}\)

\(\Rightarrow \dfrac {a}{\sin A} = \dfrac {b}{\sin B} = \dfrac {10}{1}\)

\(\Rightarrow a = 10\sin A, b = 10\sin B\)

Using equation \((1)\)

\(A(\triangle ABC) = \dfrac {1}{2} (10\sin A)(10\sin B)\)

\(= 50\sin A \sin B\)

But maximum value of \(\sin A \sin B = \dfrac {1}{2}\)

\(\therefore\) Maximum value of \(A (\triangle ABC) = 50\times \dfrac {1}{2} = 25\)

OR

\(\angle C = 90^{\circ}\Rightarrow ABC\) is right angled triangle

\(\therefore\) Area of \(\triangle\) is maximum when it is \(45^{\circ} - 45^{\circ} - 90^{\circ}\triangle\).

\(\therefore A(\triangle ABC) = \dfrac {1}{2}\times 5\sqrt {2}\times 5\sqrt {2} = 25\)


qImage671b410bd66cd79e6a2c06df

Hence option 3 is correct

Sine Rule Question 9:

If one side of a triangle is double the other and the angles opposite to these sides differ by 60°, then the triangle is

  1. obtuse angled
  2. right angled
  3. acute angled
  4. isosceles

Answer (Detailed Solution Below)

Option 2 : right angled

Sine Rule Question 9 Detailed Solution

Answer : 2

Solution :

In ΔABC, by sine rule,

\(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\)

According to the given condition,

In ΔABC, a = 2 b and

A - B = 60° A = 60° + B

⇒ \(\frac{\sin \left(60^{\circ}+B\right)}{2 b}=\frac{\sin B}{b}\)

⇒ \(\frac{\sin B}{\sin \left(B+60^{\circ}\right)}=\frac{1}{2}\)

⇒ 2 sin B = sin B cos 60° + cos B sin 60°

⇒ \(2 \sin \mathrm{~B}=\sin \mathrm{B}\left(\frac{1}{2}\right)+\cos \mathrm{B}\left(\frac{\sqrt{3}}{2}\right)\) 

⇒ \(\frac{3}{2} \sin \mathrm{~B}=\frac{\sqrt{3}}{2} \cos \mathrm{~B}\)

⇒ \(\tan B=\frac{1}{\sqrt{3}} \Rightarrow B=30^{\circ}\)

∴ A = 30° + 60° = 90°

∴ ΔABC is right angled.

Sine Rule Question 10:

In a triangle ABC, with usual notations, if c = 4, then value of \((a-b)^{2} \cos ^{2} \frac{C}{2}+(a+b)^{2} \sin ^{2} \frac{C}{2}\) is

  1. 4
  2. 16
  3. 9
  4. 25

Answer (Detailed Solution Below)

Option 2 : 16

Sine Rule Question 10 Detailed Solution

Concept Used:

1. Cosine Rule: \(c^2 = a^2 + b^2 - 2ab \cos C\)

2. Half-angle formulas:

\(\cos^2 \frac{C}{2} = \frac{1 + \cos C}{2}\)

\(\sin^2 \frac{C}{2} = \frac{1 - \cos C}{2}\)

Calculation:

Given:

In triangle ABC, c = 4

Expression: \((a - b)^2 \cos^2 \frac{C}{2} + (a + b)^2 \sin^2 \frac{C}{2}\)

\((a - b)^2 \cos^2 \frac{C}{2} + (a + b)^2 \sin^2 \frac{C}{2}\)

\((a - b)^2 \left(\frac{1 + \cos C}{2}\right) + (a + b)^2 \left(\frac{1 - \cos C}{2}\right)\)

\(\frac{1}{2} \left[ (a^2 - 2ab + b^2)(1 + \cos C) + (a^2 + 2ab + b^2)(1 - \cos C) \right]\)

\(\frac{1}{2} \left[ a^2 + a^2 \cos C - 2ab - 2ab \cos C + b^2 + b^2 \cos C + a^2 - a^2 \cos C + 2ab - 2ab \cos C + b^2 - b^2 \cos C \right]\)

\(\frac{1}{2} \left[ 2a^2 + 2b^2 - 4ab \cos C \right]\)

\(a^2 + b^2 - 2ab \cos C\)

\(c^2\) (using the cosine rule)

\(4^2\) (since c = 4)

⇒ 16

∴ The value of the expression is 16.

Hence option 2 is correct

Get Free Access Now
Hot Links: teen patti master apk teen patti refer earn teen patti casino teen patti gold download apk teen patti master download