# How do you solve the triangle given triangleABC, a=22, b=18, mangleC=130?

Jan 4, 2017

Solution is $a = 22$, $b = 18$, $c = 36.29$,

$m \angle A = {27.67}^{\circ}$, $m \angle B = {22.33}^{\circ}$ and $m \angle C = {130}^{\circ}$

#### Explanation:

Solving a triangle means finding all sides and angles of the triangle. For this we use sine formulas i.e.

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

cosine formulas ${a}^{2} = {b}^{2} + {c}^{2} - 2 b \mathcal{o} s A$, ${b}^{2} = {c}^{2} + {a}^{2} - 2 c a \cos B$ and ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$.

Here we are given $a = 22$, $b = 18$ and $m \angle C = {130}^{\circ}$, hence

${c}^{2} = {22}^{2} + {18}^{2} - 2 \times 22 \times 18 \times \cos {130}^{\circ}$

$= 484 + 324 - 792 \times \left(- 0.6428\right) = 808 + 509.0976 = 1317.0976$

i.e. $c = 36.29$ and using sine formula, we get

$\frac{22}{\sin} A = \frac{18}{\sin} B = \frac{36.29}{\sin {130}^{\circ}} = \frac{36.29}{0.766} = 47.376$

Hence, $\sin A = \frac{22}{47.376} = 0.4644$ and $A = {27.67}^{\circ}$

and $\sin B = \frac{18}{47.376} = 0.3799$ and $B = {22.33}^{\circ}$

Solution is $a = 22$, $b = 18$, $c = 36.29$,

$m \angle A = {27.67}^{\circ}$, $m \angle B = {22.33}^{\circ}$ and $m \angle C = {130}^{\circ}$