NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

The solution of $\Delta ABC$ given that $B = 45^\circ$ , $C = 105^\circ$ and $c = \sqrt{2}$ is:

Choose the correct answer:

Explanation

1. Find Angle $A$ :

The sum of angles in a triangle is $180^\circ$ .

A + B + C = 180^\circ

A + 45^\circ + 105^\circ = 180^\circ

A = 180^\circ - 150^\circ

A = 30^\circ

2. Find Side $b$ using Sine Rule:

The Sine Rule states: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Using $\frac{b}{\sin B} = \frac{c}{\sin C}$ :

\frac{b}{\sin 45^\circ} = \frac{\sqrt{2}}{\sin 105^\circ}

Note: $\sin 105^\circ = \sin(60^\circ + 45^\circ) = \frac{\sqrt{3}+1}{2\sqrt{2}}$

b = \frac{\sqrt{2} \times \sin 45^\circ}{\sin 105^\circ} = \frac{\sqrt{2} \times \frac{1}{\sqrt{2}}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}

b = \frac{2\sqrt{2}}{\sqrt{3}+1}

Rationalizing the denominator:

b = \frac{2\sqrt{2}(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{2\sqrt{2}(\sqrt{3}-1)}{3-1} = \frac{2\sqrt{2}(\sqrt{3}-1)}{2}

b = \sqrt{2}(\sqrt{3}-1)

3. Find Side $a$ using Sine Rule:

Using $\frac{a}{\sin A} = \frac{c}{\sin C}$ :

\frac{a}{\sin 30^\circ} = \frac{\sqrt{2}}{\sin 105^\circ}

a = \frac{\sqrt{2} \times \frac{1}{2}}{\frac{\sqrt{3}+1}{2\sqrt{2}}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}+1}{2\sqrt{2}}}

a = \frac{\sqrt{2}}{2} \times \frac{2\sqrt{2}}{\sqrt{3}+1} = \frac{2}{\sqrt{3}+1}

Rationalizing the denominator:

a = \frac{2(\sqrt{3}-1)}{3-1} = \frac{2(\sqrt{3}-1)}{2}

a = \sqrt{3}-1

Conclusion:

The values are $A = 30^\circ$ , $a = \sqrt{3}-1$ , and $b = \sqrt{2}(\sqrt{3}-1)$ .

The correct option is (a).