Question

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

Answer 1

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^2-2bc*cos(alpha)`

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

Using the formula above,
`16=49+64-2*7*8*cos(alpha)`
from which
`cos(alpha)=97/112=0.86607143` (approximately)
The angle, whose cosine approximately equals to `0.86607143` is about `30^0` since `cos(30^0)=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.