Question

In a $\triangle ABC$, b = 2, C = $60^{\circ}$, c =$\sqrt{6}$ , then a =
A. $\sqrt{3}-1$
B. $\sqrt{3}$
C. $\sqrt{3}+1$
D. None of these

Answer 1

Hint: We have a $\triangle ABC$ with b = 2, C = $60^{\circ}$, c =$\sqrt{6}$ . We are asked to find the value for a. Recall the law of sines.

Formula Used:Law of sines: For a $\triangle ABC$,
$\dfrac{sin A}{a}=\dfrac{sin B}{b}=\dfrac{sin C}{c}$

Complete Step-by-Step : We have a with b = 2, C =$60^{\circ}$, c =$\sqrt{6}$. We have to find the value of a.
We have,
$\dfrac{sin A}{a}=\dfrac{sin B}{b}=\dfrac{sin C}{c}$
We given, b = 2, C =$60^{\circ}$, c =$\sqrt{6}$, substituting these values in the above equation we get,
$\dfrac{sin A}{a} = \dfrac{sin B}{2} = \dfrac{sin 60^{\circ}}{\sqrt{6}}$
Let us consider the ratio,
$\dfrac{sin B}{2} = \dfrac{sin 60^{\circ}}{\sqrt{6}}$
$\implies sin B = \dfrac{2\times sin 60^{\circ}}{\sqrt{6}}$
$\implies sin B = \dfrac{2\times \sqrt{3}}{\sqrt{6}\times 2}$
$\implies sin B = \dfrac{\sqrt{3}}{\sqrt{6}}$
$\implies sin B = \dfrac{\sqrt{3}}{\sqrt{3}\times\sqrt{2}}$
$\therefore sin B = \dfrac{1}{\sqrt{2}}$
$\implies B = sin^{-1}(\dfrac{1}{\sqrt{2}})$
$\implies B = 45^{\circ}$
Now, we consider the first two ratios.
That is, $\dfrac{sin A}{a} = \dfrac{sin B}{2}$
Therefore, we get
$\dfrac{sin A}{a} = \dfrac{\dfrac{1}{\sqrt{2}}}{2}$
$\implies \dfrac{sin A}{a} =\dfrac{1}{2\sqrt{2}}$
To find the value of A, we know that the sum of angle of triangle is 180.
$\therefore A+B+C = 180^{\circ}$
$\implies A + 45^{\circ}+60^{\circ} =180^{\circ}$
$\implies A + 105^{\circ} =180^{\circ}$
$\implies A =180^{\circ}-105^{\circ}$
$\therefore A = 75^{\circ}$
Substituting A =$75^{\circ}$ , in we get,
$\dfrac{sin 75^{\circ}}{a} =\dfrac{1}{2\sqrt{2}}$
$\implies a = 2\sqrt{2}\times sin 75^{\circ}$
Now, we need to find the value of $sin 75^{\circ}$.
We can write $sin 75^{\circ}$ as $sin (30^{\circ}+45^{\circ})$
$\therefore sin 75^{\circ} =sin (30^{\circ}+45^{\circ})$
We know that $sin(x+y)=sinxcosy+sinycosx$
$\therefore sin 75^{\circ} =sin (30^{\circ}+45^{\circ})=sin30^{\circ}cos45^{\circ}+sin45^{\circ}cos30^{\circ}$
$sin 75^{\circ} = \dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}$
$sin 75^{\circ}=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}$
$\therefore sin 75^{\circ}=\dfrac{1+\sqrt{3}}{2\sqrt{2}}$
We have , $a = 2\sqrt{2}\times sin 75^{\circ}$
$\implies a =2\sqrt{2}\times\dfrac{1+\sqrt{3}}{2\sqrt{2}}$
$\therefore a = 1+\sqrt{3}$
Therefore, the value of a is $1+\sqrt{3}$. So, the answer is Option C.

Note: In this question, we are given two sides and an angle. Therefore, it is possible to solve this question by using the law of cosines. We have,$c^{2} = a^{2}+b^{2}-2ab~cosC$. Rearranging the equation we get, $cos C = \dfrac{a^{2}+b^{2}-c^{2}}{2ab}$. Substituting the value we get a quadratic equation in a, by solving this equation we get the value for A.