Question

Find the degree measure of the smallest angle in a triangle with sides 8, 10, and 12.

Answer 1

Let $\angle ABC$ be the smallest angle of the triangle below.

[asy]
pair A,B,C;
A=(0,0);
B=(4,0);
C=(1.2,2.4);
draw(A--B--C--A);
draw(rightanglemark(B,A,C,4.5));
label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,N);
[/asy]

Since the sum of the angles in a triangle is $180^\circ$, and $\angle ABC$ is the smallest angle, we have \begin{align*}
\angle ABC + \angle BCA + \angle CAB & = 180^\circ \\
\angle ABC + \angle BCA + \angle CAB & = \angle ABC + (180^\circ - \angle ABC - \angle CAB) + \angle CAB = 180^\circ.
\end{align*} Simplifying the second equation gives $\angle ABC = \boxed{30^\circ}$.

Answer 2

Find the degree measure of the smallest angle in a triangle with sides of 8cm, 10cm, and 12cm.

Answer 3

Let $a$, $b$, and $c$ be the lengths of the sides of the triangle, and let $A$, $B$, and $C$ be the measures of the angles opposite those sides, respectively.

By the Law of Cosines, we have
\[c^2 = a^2 + b^2 - 2ab \cos C.\]Since 12 is the longest side of the triangle, we have $12^2 = 8^2 + 10^2 - 2(8)(10) \cos C$, or $144 = 164 - 160 \cos C.$ Solving, we find $\cos C = \frac{5}{8},$ so
\[C \approx 37^\circ.\]Since this is the largest angle of the triangle, the smallest angle measures $\boxed{37^\circ}.$