# The lengths of the sides of a triangle are 7, 8 and 9 cm. How do you calculate the size of the smallest angle in the triangle?

Jun 23, 2016

Use the Law of Cosines (and a calculator) to determine:
smallest angle $\approx 0.841069$ radians

#### Explanation:

Denote the sides as
$\textcolor{w h i t e}{\text{XXX}} a = 9$
$\textcolor{w h i t e}{\text{XXX}} b = 8$
$\textcolor{w h i t e}{\text{XXX}} c = 7$
and
$\textcolor{w h i t e}{\text{XXX}}$the angle opposite $a$ as $A$
$\textcolor{w h i t e}{\text{XXX}}$the angle opposite $b$ as $B$
$\textcolor{w h i t e}{\text{XXX}}$the angle opposite $c$ as $C$

Note that the smallest angle is the angle opposite the shortest side (i.e. angle $C$ in this case).

By The Law of Cosines:
$\textcolor{w h i t e}{\text{XXX}} {c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cdot \cos \left(C\right)$
or
$\textcolor{w h i t e}{\text{XXX}} \cos \left(C\right) = \frac{\left({a}^{2} + {b}^{2}\right) - {c}^{2}}{2} a b$

Using the given values
$\textcolor{w h i t e}{\text{XXX}} \cos \left(C\right) = \frac{{8}^{2} + {9}^{2} - {7}^{2}}{2 \cdot 8 \cdot 9} = \frac{96}{144} = \frac{2}{3}$

If $\cos \left(C\right) = \frac{2}{3}$ then
$\textcolor{w h i t e}{\text{XXX}} C = \arccos \left(\frac{2}{3}\right)$ (here is where I used a calculator)
$\textcolor{w h i t e}{\text{XXXX}} \approx 0.841069$ radians