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

Calculation

Let the sides be \(a-d, a, a+d\).

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

By given condition, \(C = 2A\).

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

Hence by sine rule we have,

\(\dfrac{a+d}{\sin C} = \dfrac{a-d}{\sin A} = \dfrac{a}{\sin B}\)

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

\(\therefore 2\cos A = \dfrac{a+d}{a-d}\) and \(\dfrac{a}{a-d} = \dfrac{\sin 3A}{\sin A}\).

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

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

\(a \neq 0\) \(\therefore a-d = 4d\) 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, \(\cos A = \dfrac{AC}{AB}\).

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

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

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.

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

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

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,

\(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = {\rm{\;}}\frac{{\rm{b}}}{{\sin {\rm{B}}}} \)

\(\rm ⇒ \frac{b}{a}=\frac{\sin B}{\sin A}\)

\(\rm ⇒ \frac{√3 }{1}=\frac{\sin 60^o}{\sin A}\)

⇒ sin A = (√3/2) × (1/√3)

⇒ sin A = 1/2

⇒ A = 30° 

Hence, option (1) is correct. 

Concept:

Sine Rule:

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

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

\(\rm \frac{sin\: A}{a} = \frac{sin\: B}{b}\)

⇒ \(\rm \frac{1/4}{2} = \frac{sin\: B}{4}\)

⇒ sin B = \(\rm 1\over2\)

⇒ B = 30∘ = \(\rm \pi\over6\)

Additional Information

Cosine Rule:

cos A = \(\rm \frac{b^{2} + c^{2} - a^{2}}{2bc}\)

cos B = \(\rm \frac{c^{2} + a^{2} - b^{2}}{2ac}\)

cos C = \(\rm \frac{a^{2} + b^{2} - c^{2}}{2ab}\)

Concept:

In ΔABC, sine rule

\(\rm \dfrac {a}{sin\; A} = \dfrac {b}{sin\; B} = \dfrac {c}{sin\;C} \)

\(\rm A (ΔABC) = \dfrac{1}{2}\times base \;\times height \)

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.

\(\rm \angle C = 90^\circ\)

Hence,  ΔABC is right angled triangle.

\(\rm A (ΔABC) = \dfrac{1}{2}\times base \;\times height \)

\(\rm A (ΔABC) = \dfrac{1}{2}\times a \times b\)....(1)

By sine rule, we have

\(\rm \dfrac {a}{sin\; A} = \dfrac {b}{sin\; B} = \dfrac {c}{sin\;C} \)

⇒\(\rm \dfrac {a}{sin\; A} = \dfrac {b}{sin\; B} = \dfrac {10}{sin\; 90^\circ} \)

⇒\(\rm \dfrac {a}{sin\; A} = \dfrac {b}{sin\; B} = \dfrac {10}{1} \)

⇒\(\rm a = 10\;sin\;A\) and \(\rm b = 10\;sin\;B\)

 Equation (1) becomes, 

\(\rm A (ΔABC) = \dfrac{1}{2}\times 10 \;sin\; A \times 10\;sin\; B \)

⇒\(\rm A (ΔABC) = 50 \times sin\; A \times sin\; B \) ....(2)

Since, \(\rm \angle C = 90^\circ\)and sum of all angles of triangle is 180º

\(\rm \angle (A+B) = 90^\circ\)

⇒Max (\(\rm sin\; A \times sin\; B\)) = \(\rm \dfrac 1 2\)

Equation (2) becomes,

\(\rm A (ΔABC) = 50 \times \dfrac 1 2 \)

\(\rm A (ΔABC) = 25 \)

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:

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}\)

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

Calculation:

1. We have,

sin \(\rm B = \frac 1 3=\frac{P}{H}\)

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,

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}\)

⇒ 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°, \({\rm{a}} = 2\sqrt 2 \),

Using sine rule in given triangle,

 \(\frac{a}{\text{sinA}}=\frac{c}{\text{sinC}}\\ \Rightarrow \frac{2\sqrt2}{\text{sin45}}=\frac{2}{\text{sinC}}\\\Rightarrow \text{sinC} = \frac{\text{sin45}}{\sqrt2}=\frac{1}{\sqrt2\sqrt2}\\ \Rightarrow \text{sinC} =\frac{1}{2}\\ \Rightarrow C =30\)

Hence, option (1) is correct.

Concept:

Concept:

In triangle ABC, According to the Sine rule

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

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

\(\frac{a}{\text{sinA}}=\frac{b}{\text{sinB}}\)

⇒ \(\rm \frac {√3}{sin\ 60^0}\ = \ \frac {b}{sin\ 45^0}\)

⇒ \(\rm \frac {√3}{\frac {√3}{2}}\ = \ \frac {b}{\frac {1}{√2}}\)

⇒ \(\rm b\ = \frac {2}{√2}\)

⇒ b = √2 

Concept:

Consider a triangle ABC,

Cosine rule:

\(\cos {\rm{C}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}\)

Sine rule:

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

Calculation:

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

We know that, \(\cos {\rm{C}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}\)

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

⇒ 2(1 + √3) = 1 + 3 + 2√3 + 4 – c²

⇒ c² = 6

⇒ c = √6

Using sin rule we get,

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

\( \Rightarrow \frac{{1 + \sqrt 3 }}{{\sin {\rm{A}}}} = \frac{2}{{\sin {\rm{B}}}} = \frac{{\sqrt 6 }}{{\sin 60^\circ }}\)

⇒ sin B = (2 / √6) × sin 60°

\( \Rightarrow \sin {\rm{B}} = \frac{1}{{\sqrt 2 }}\)

⇒ B = 45°

Now, sum of all angles of triangles = 180°

⇒ A + B + C = 180°

⇒ A = 180° – 45° – 60°

⇒ A = 75°

Concept:

Sine Rule:

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

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

\(\rm \frac{sin\: A}{a} = \frac{sin\: B}{b}\)

\(\rm \frac{3/4}{3} = \frac{sin\: B}{4}\)

sin B = 1

B = 90∘ = π/2

Additional Information

Cosine Rule:

cos A = \(\rm \frac{b^{2} + c^{2} - a^{2}}{2bc}\)

cos B = \(\rm \frac{c^{2} + a^{2} - b^{2}}{2ac}\)

cos C = \(\rm \frac{a^{2} + b^{2} - c^{2}}{2ab}\)

Concept:

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

Calculations:

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

We know that, \(\rm \frac{a}{\ sin\;A}=\frac{b}{sin\;B}=\frac{c}{sin\;C} =\rm k\)

⇒\(\rm \ sin \;A = \frac{a}{k}, \ sin\;B = \frac{b}{k}, \ sin\;C = {c}{k}\)

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

  \(\rm \frac{a}{\ sin\;30}=\frac{b}{sin\;60}=\frac{c}{sin\;90} =\rm k\)

  \(\rm \frac{a}{\frac{1}{2}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{c}{1} =\rm k\)

  \(\rm \frac{a}{1}=\frac{b}{\sqrt{3}}=\frac{c}{2} =\rm k\) 

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