# How do you find the measure of the smallest in an acute triangle whose side lengths are 4m, 7m, and 8m?

Apr 20, 2015

The smallest angle lies across the smallest side, in this case it's the one that measures 4m.

Now recall the Law of Cosines:
${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cdot \cos \left(\alpha\right)$

In our case
$a = 4 m$,
$b = 7 m$,
$c = 8 m$ and
$\alpha$ - is the smallest angle we need to measure.

Using the formula above,
$16 = 49 + 64 - 2 \cdot 7 \cdot 8 \cdot \cos \left(\alpha\right)$
from which
$\cos \left(\alpha\right) = \frac{97}{112} = 0.86607143$ (approximately)
The angle, whose cosine approximately equals to $0.86607143$ is about ${30}^{0}$ since $\cos \left({30}^{0}\right) = 0.86602540$ (approximately).

More information on the Law of Cosine you can find at Unizor by following the menu items Trigonometry - Simple Trigonometric Identities - Law of Cosine.