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

Explanation:

We are given the triangle with angles:

$\mathrm{\angle }B=x={30}^{\circ }$

$\mathrm{\angle }A=3x={90}^{\circ }$

$\mathrm{\angle }C={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

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

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={90}^{\circ }+{30}^{\circ }+{60}^{\circ }={180}^{\circ }$

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

Statement I. The triangle is right-angled.

Since$\mathrm{\angle }A={90}^{\circ }$, the triangle is right-angled.

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

The angles ${30}^{\circ }{60}^{\circ },\text{a}nd{90}^{\circ }$ are in Arithmetic Progression because:

${60}^{\circ }-{30}^{\circ }={30}^{\circ }\text{and}{90}^{\circ }-{60}^{\circ }={30}^{\circ }$

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.

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:

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

$x={30}^{\circ }$

∴ The correct answer is Option (2)

Calculation

Let the sides be $a-d,a,a+d$.

It is understood that $a>d>0$ and from the figure, $\mathrm{\angle }C$ is greatest and $\mathrm{\angle }A$ is smallest.

By given condition, $C=2A$.

And hence $B=\pi -(A+C)=\pi -3A$.

Hence by sine rule we have,

${\displaystyle \frac{a+d}{\sin C}}={\displaystyle \frac{a-d}{\sin A}}={\displaystyle \frac{a}{\sin B}}$

or ${\displaystyle \frac{a+d}{\sin 2A}}={\displaystyle \frac{a-d}{\sin A}}={\displaystyle \frac{a}{\sin (\pi -3A)}}$.

$\therefore 2\cos A={\displaystyle \frac{a+d}{a-d}}$ and ${\displaystyle \frac{a}{a-d}}={\displaystyle \frac{\sin 3A}{\sin A}}$.

$\therefore {\displaystyle \frac{a}{a-d}}=3-4{\sin }^{2}A=3-4+(2\cos A{)}^2$.

$\therefore {\displaystyle \frac{a}{a-d}}=1+{\left({\displaystyle \frac{a+d}{a-d}}\right)}^2={\displaystyle \frac{4ad}{(a-d{)}^2}}$.

$a\ne 0$ $\therefore a-d=4d$ or $a=5d$.

$\therefore$ sides are $a-d,a,a+d$

or $4d,5d,6d$.

Hence the required ratio is $4:5:6$.

Then the sum of ratio of its sides is  = 15

Thus, $\cos A={\displaystyle \frac{AC}{AB}}$.

Using above relation we can claim that $\mathrm{△}ABC$ is a right angled triangle with AB as hypotenuse, right angled at C.

Calculation

${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C$

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

$A(\mathrm{△}ABC)={\displaystyle \frac{1}{2}}ab.....(1)$

From sine rule ${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}$

$\Rightarrow {\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{10}{1}}$

$\Rightarrow a=10\sin A,b=10\sin B$

Using equation $(1)$

$A(\mathrm{△}ABC)={\displaystyle \frac{1}{2}}(10\sin A)(10\sin B)$

$=50\sin A\sin B$

But maximum value of $\sin A\sin B={\displaystyle \frac{1}{2}}$

$\therefore$ Maximum value of $A(\mathrm{△}ABC)=50\times {\displaystyle \frac{1}{2}}=25$

OR

$\mathrm{\angle }C={90}^{\circ }\Rightarrow ABC$ is right angled triangle

$\therefore$ Area of $\mathrm{△}$ is maximum when it is ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }\mathrm{△}$.

$\therefore A(\mathrm{△}ABC)={\displaystyle \frac{1}{2}}\times 5\sqrt{2}\times 5\sqrt{2}=25$

Hence option 3 is correct

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.

Concepts:

Sine law ⇔ $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$

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

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,

$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}$

$\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\sin \mathrm{A}}$

$\Rightarrow \frac{\surd 3}{1}=\frac{\sin {60}^{\mathrm{o}}}{\sin \mathrm{A}}$

⇒ sin A = (√3/2) × (1/√3)

⇒ sin A = 1/2

⇒ A = 30° 

Hence, option (1) is correct. 

Concept:

Sine Rule:

$\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}{\mathrm{a}}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}{\mathrm{b}}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}}{\mathrm{c}}$

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

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

$\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}{\mathrm{a}}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}{\mathrm{b}}$

⇒ $\frac{1/4}{2}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}{4}$

⇒ sin B = $\frac{1}{2}$

⇒ B = 30∘ = $\frac{\pi }{6}$

Additional Information

Cosine Rule:

cos A = $\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{b}\mathrm{c}}$

cos B = $\frac{{\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2}{2\mathrm{a}\mathrm{c}}$

cos C = $\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{a}\mathrm{b}}$

Concept:

In ΔABC, sine rule

${\displaystyle \frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}}={\displaystyle \frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}}={\displaystyle \frac{\mathrm{c}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}}}$

$\mathrm{A}(\Delta \mathrm{A}\mathrm{B}\mathrm{C})={\displaystyle \frac{1}{2}}\times \mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}\times \mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}$

Calculations:

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

length of Side (AB) = 10 ⇒ c = 10

⇒ a2 + b2 = c2

Pythagoreans theorem is verified.

$\mathrm{\angle }\mathrm{C}={90}^{\circ }$

Hence,  ΔABC is right angled triangle.

$\mathrm{A}(\Delta \mathrm{A}\mathrm{B}\mathrm{C})={\displaystyle \frac{1}{2}}\times \mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}\times \mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}$

$\mathrm{A}(\Delta \mathrm{A}\mathrm{B}\mathrm{C})={\displaystyle \frac{1}{2}}\times \mathrm{a}\times \mathrm{b}$....(1)

By sine rule, we have

${\displaystyle \frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}}={\displaystyle \frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}}={\displaystyle \frac{\mathrm{c}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}}}$

⇒${\displaystyle \frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}}={\displaystyle \frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}}={\displaystyle \frac{10}{\mathrm{s}\mathrm{i}\mathrm{n}{90}^{\circ }}}$

⇒${\displaystyle \frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}}={\displaystyle \frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}}={\displaystyle \frac{10}{1}}$

⇒$\mathrm{a}=10\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}$ and $\mathrm{b}=10\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}$

 Equation (1) becomes, 

$\mathrm{A}(\Delta \mathrm{A}\mathrm{B}\mathrm{C})={\displaystyle \frac{1}{2}}\times 10\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}\times 10\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}$

⇒$\mathrm{A}(\Delta \mathrm{A}\mathrm{B}\mathrm{C})=50\times \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}\times \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}$ ....(2)

Since, $\mathrm{\angle }\mathrm{C}={90}^{\circ }$and sum of all angles of triangle is 180º

$\mathrm{\angle }(\mathrm{A}+\mathrm{B})={90}^{\circ }$

⇒Max ($\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}\times \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}$) = ${\displaystyle \frac{1}{2}}$

Equation (2) becomes,

$\mathrm{A}(\Delta \mathrm{A}\mathrm{B}\mathrm{C})=50\times {\displaystyle \frac{1}{2}}$

$\mathrm{A}(\Delta \mathrm{A}\mathrm{B}\mathrm{C})=25$

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:

$\frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}}=\frac{1}{\mathrm{K}}$

sin A - cos B = $2\cos (\frac{\mathrm{A}+\mathrm{B}}{2})\sin (\frac{\mathrm{A}-\mathrm{B}}{2})$

Calculation:

1. We have,

sin $\mathrm{B}=\frac{1}{3}=\frac{\mathrm{P}}{\mathrm{H}}$

Hwence, Right angle triangle can be drawn as,

Here, P = 1
H = 3
B = x

Using Pythagoras theorem,

3² = 1² + x²

x² = 8

$\therefore x=2\sqrt{2}$

Now , Cosec C = $\frac{3}{2\sqrt{2}}$

Hence Ststement (1) is not correct.

2. Given,

b cos B = c cos C

Using the sine rule,

$\frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}}=\frac{1}{\mathrm{K}}$

⇒ 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 $(\frac{C+D}{2})$ × sin $(\frac{C-D}{2})$

⇒ 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.

Concept:

In triangle ABC, According to Sine rule

$\frac{a}{\text{sinA}}=\frac{b}{\text{sinB}}=\frac{c}{\text{sinC}}$ Where, a, b, c are sides and A, B, C are angles of triangle.

Calculation:

In a triangle ABC, c = 2, A = 45°, $\mathrm{a}=2\sqrt{2}$,

Using sine rule in given triangle,

 $$\begin{gathered} \frac{a}{\text{sinA}}=\frac{c}{\text{sinC}} \\ \Rightarrow \frac{2\sqrt{2}}{\text{sin45}}=\frac{2}{\text{sinC}} \\ \Rightarrow \text{sinC}=\frac{\text{sin45}}{\sqrt{2}}=\frac{1}{\sqrt{2}\sqrt{2}} \\ \Rightarrow \text{sinC}=\frac{1}{2} \\ \Rightarrow C=30 \end{gathered}$$

Hence, option (1) is correct.

Concept:

Concept:

In triangle ABC, According to the Sine rule

$\frac{\mathrm{a}}{\text{sin A}}=\frac{\mathrm{b}}{\text{sin B}}=\frac{\mathrm{c}}{\text{sin C}}$

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

$\frac{a}{\text{sinA}}=\frac{b}{\text{sinB}}$

⇒ $\frac{\surd 3}{\mathrm{s}\mathrm{i}\mathrm{n}{60}^0}=\frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}{45}^0}$

⇒ $\frac{\surd 3}{\frac{\surd 3}{2}}=\frac{\mathrm{b}}{\frac{1}{\surd 2}}$

⇒ $\mathrm{b}=\frac{2}{\surd 2}$

⇒ b = √2 

Concept:

Consider a triangle ABC,

Cosine rule:

$\cos \mathrm{C}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{a}\mathrm{b}}$

Sine rule:

$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$

Calculation:

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

We know that, $\cos \mathrm{C}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{a}\mathrm{b}}$

$\Rightarrow \cos {60}^{\circ }=\frac{{\left(1+\sqrt{3}\right)}^2+2^2-{\mathrm{c}}^2}{2\left(1+\sqrt{3}\right)\left(2\right)}$

⇒ 2(1 + √3) = 1 + 3 + 2√3 + 4 – c²

⇒ c² = 6

⇒ c = √6

Using sin rule we get,

$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$

$\Rightarrow \frac{1+\sqrt{3}}{\sin \mathrm{A}}=\frac{2}{\sin \mathrm{B}}=\frac{\sqrt{6}}{\sin {60}^{\circ }}$

⇒ sin B = (2 / √6) × sin 60°

$\Rightarrow \sin \mathrm{B}=\frac{1}{\sqrt{2}}$

⇒ B = 45°

Now, sum of all angles of triangles = 180°

⇒ A + B + C = 180°

⇒ A = 180° – 45° – 60°

⇒ A = 75°

Concept:

Sine Rule:

$\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}{\mathrm{a}}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}{\mathrm{b}}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}}{\mathrm{c}}$

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

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

$\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}{\mathrm{a}}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}{\mathrm{b}}$

$\frac{3/4}{3}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}{4}$

sin B = 1

B = 90∘ = π/2

Additional Information

Cosine Rule:

cos A = $\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{b}\mathrm{c}}$

cos B = $\frac{{\mathrm{c}}^2+{\mathrm{a}}^2-{\mathrm{b}}^2}{2\mathrm{a}\mathrm{c}}$

cos C = $\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{a}\mathrm{b}}$

Concept:

$\frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}}=\mathrm{k}$

Calculations:

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

We know that, $\frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}}=\mathrm{k}$

⇒$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}=\frac{\mathrm{a}}{\mathrm{k}},\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}=\frac{\mathrm{b}}{\mathrm{k}},\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}=\mathrm{c}\mathrm{k}$

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, $\frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}=\frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{B}}=\frac{\mathrm{c}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}}=\mathrm{k}$

  $\frac{\mathrm{a}}{\mathrm{s}\mathrm{i}\mathrm{n}30}=\frac{\mathrm{b}}{\mathrm{s}\mathrm{i}\mathrm{n}60}=\frac{\mathrm{c}}{\mathrm{s}\mathrm{i}\mathrm{n}90}=\mathrm{k}$

  $\frac{\mathrm{a}}{\frac{1}{2}}=\frac{\mathrm{b}}{\frac{\sqrt{3}}{2}}=\frac{\mathrm{c}}{1}=\mathrm{k}$

  $\frac{\mathrm{a}}{1}=\frac{\mathrm{b}}{\sqrt{3}}=\frac{\mathrm{c}}{2}=\mathrm{k}$ 

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