# How do you solve the triangle ABC given C=84^circ, c=7, a=2?

Feb 24, 2018

$a = 2 , b = 6.92 , c = 7$

$\hat{A} = {16.5}^{\circ} , \hat{B} = {79.5}^{\circ} , \hat{C} = {84}^{\circ}$

#### Explanation:

Given : $\hat{C} = {84}^{\circ} , a = 2 , c = 7$

Applying law of sines,

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

$\sin A = \frac{a \sin C}{\sin} A = \left(2 \cdot \sin 84\right) / 7 = 0.2841$

$\hat{A} = {\sin}^{-} 1 \left(0.2841\right) \approx {16.5}^{\circ}$

Sum of the three angles of a triangle equals ${180}^{\circ}$

Hence $\hat{B} = 180 - 84 - 16.5 = {79.5}^{\circ}$

$b = \frac{c \sin B}{\sin} C = \frac{7 \cdot \sin 79.5}{\sin} 84 = 6.92$