Question

In a triangle ABC, if a = 2, B =$60^{\circ}$ and C =$75^{\circ}$ , then b =
A. $\sqrt{3}$
B.$\sqrt{6}$
C. $\sqrt{9}$
D. $1+\sqrt{2}$

Answer 1

Hint: We are given a triangle ABC with a = 2, B =$60^{\circ}$ and C =$75^{\circ}$ . We need to find the measure of side b. Properties of a triangle and the law of sines.
Formula used:
From the law of sines we have,
$\dfrac{sin A}{a} = \dfrac{sin B}{b} = \dfrac{sin C}{c}$
Complete Step-by-Step : We have triangle ABC and a = 2, B = $60^{\circ}$and C =$75^{\circ}$ .

Since Sum of Angle of triangle is 180, we get,
$\angle A+\angle B+\angle C = 180^{\circ}$
$\angle A + 60^{\circ} + 75^{\circ} = 180^{\circ}$
$\angle A = 180^{\circ}-(60^{\circ} + 75^{\circ})$
$\angle A = 180^{\circ}-135^{\circ}$
$\angle A = 45^{\circ}$
From the law of sines we have,
$\dfrac{sin A}{a} = \dfrac{sin B}{b} = \dfrac{sin C}{c}$
We can find the value of b by using the first two ratios,
$\dfrac{sin A}{a} = \dfrac{sin B}{b}$
Substituting the values of a, and we get,
$\dfrac{sin 45^{\circ}}{2} = \dfrac{sin 60^{\circ}}{b}$
$\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2} = \dfrac{\sqrt{3}}{2}\times\dfrac{1}{b}$
$\dfrac{1}{2\sqrt{2}} = \dfrac{\sqrt{3}}{2b}$
$\implies 2b = \sqrt{3}\times2\sqrt{2}$
$\implies b = \sqrt{3}\times\sqrt{2}$
$\implies b = \sqrt{6}$
Therefore, the measure of b is $\sqrt{6}$. So, the answer is Option B.
Note:To solve this question, we have used the law of sines. You may have thought of using the tan ratio for solving the same, but it does not work since the tan ratio works only with the right triangles and the triangle ABC that we have in the question is not a right triangle. It is due to this reason that we used the law of sines to find the value of b.