Sine Rule MCQ Quiz in मल्याळम - Objective Question with Answer for Sine Rule - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 12, 2025
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
Answer (Detailed Solution Below)
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:
- cos(π - θ) = -cos θ
- 1 - sin2θ = cos2θ
- cos(A + B) cos(A − B) = cos2A - sin2Bcos(A+B)cos(A−B)=cos2A−sin2B
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 -
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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:
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?
Answer (Detailed Solution Below)
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:
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
Answer (Detailed Solution Below)
Sine Rule Question 5 Detailed Solution
Calculation:
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:
Answer (Detailed Solution Below)
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 _______.
Answer (Detailed Solution Below)
Sine Rule Question 7 Detailed Solution
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
Answer (Detailed Solution Below)
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 \(45^{\circ} - 45^{\circ} - 90^{\circ}\triangle\).
\(\therefore A(\triangle ABC) = \dfrac {1}{2}\times 5\sqrt {2}\times 5\sqrt {2} = 25\)
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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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