# Question #a69e0

Nov 18, 2016

$A \approx {44.2}^{\circ}$
$B \approx {95.5}^{\circ}$
$C \approx {40.3}^{\circ}$

#### Explanation:

By convention, $a$ represents the side opposite vertex $A$, $b$ represents the side opposite vertex $B$, and $c$ represents the side opposite vertex $C$.

The law of cosines states that ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(C\right)$. Note that this is true regardless of how the sides and vertices are labeled, so it is also true that

• ${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cos \left(B\right)$
• ${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos \left(A\right)$

If we try solving for the angle, we get

• $\cos \left(A\right) = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}$
• $\cos \left(B\right) = \frac{{a}^{2} + {c}^{2} - {b}^{2}}{2 a c}$
• $\cos \left(C\right) = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}$

Thus, applying the inverse cosine function to both sides:

• $A = \arccos \left(\frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}\right)$
• $B = \arccos \left(\frac{{a}^{2} + {c}^{2} - {b}^{2}}{2 a c}\right)$
• $C = \arccos \left(\frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}\right)$

Substituting in the given values $a = 14 , b = 20 , c = 13$ into a calculator, we get our results:

• $A \approx {44.2}^{\circ}$
• $B \approx {95.5}^{\circ}$
• $C \approx {40.3}^{\circ}$