# How do you solve a triangle if you are given a=2.5, b=10.2, c=9?

Jul 8, 2015

$\angle A = 0.229406$ radians
$\angle B = 1.953156$ radians
$\angle C = 0.959031$ radians

#### Explanation:

For a triangle with
$\textcolor{w h i t e}{\text{XXXX}}$side $a$ opposite angle $A$
$\textcolor{w h i t e}{\text{XXXX}}$side $b$ opposite angle $B$
$\textcolor{w h i t e}{\text{XXXX}}$side $c$ opposite angle $C$
The Law of Cosines says:
$\textcolor{w h i t e}{\text{XXXX}}$${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(C\right)$
or
$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left(C\right) = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(and similarly for $A$ and $B$)

So
$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left(C\right) = \frac{{2.5}^{2} + {10.2}^{2} - {9}^{2}}{2 \left(2.5\right) \left(10.2\right)}$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= 0.574314$

$C = \text{arccos} \left(\cos \left(C\right)\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$C = \text{arccos} \left(0.574314\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= 0.959031$ radians
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(using a calculator)

Similar calculations can be made for angles $A$ and $B$