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

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(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

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

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

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

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 

\(a < b < c\) 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 \(= 4 : 5 : 6\)

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

Calculation

\(\frac{c }{sin 30} = \frac{4 √ 3}{ sin 120}\)   [By sine rule]

⇒ 2c = 8 ⇒ c = 4

⇒ AB = |(b + 1)| = 4

⇒ b = 3 ,   m_AB = 0 

⇒ m_BC =\( \frac{− 1 }{√ 3 }\) 

BC : − y = \(\frac{−1}{√ 3}( x − 3 ) \)

⇒ √ 3y + x = 3

Point of intersection:  y = x + 3 , √ 3y + x = 3

⇒ \((\sqrt{3} + 1) y = 6 \)

⇒ \(y = \frac{6}{ √ 3 + 1}\)

⇒ \(x = \frac{6} {√ 3 + 1 }− 3 \)

⇒ x =\(\frac{ − 6}{ ( 1 + √ 3 )^ 2}\)

\(\rm \frac{\beta^2}{\alpha^2}\) = \(\frac{6^4}{ (√ 3 + 1)^4}\times\frac{ ( 1 + √ 3 )^ 4}{ (− 6)^2}\) = 36

Given :

△ABC in which ∠B=∠C.

To prove :  AB=AC

Construction :
Draw the bisector of ∠A and let it meet BC at D.
Proof  :

In △ ABD and ACD,

∠B=∠C  (Given)

∠BAD=∠CAD

AD=AD (Common)

So, by AAS criterion of congruence,

△ABD≅△ACD

⟹ AB=AC (BY Corresponding Part of Congruent Triangle)