What is the radius of the circle inscribed in triangle $ABC$ if $AB = 12,$ $AC=14,$ $BC=20$? Express your answer in simplest radical form.

Answer 1

Using the Law of Cosines, $\cos B = \frac{20^2 - 12^2 - 14^2}{2 \cdot 12 \cdot 14} = -\frac{1}{7}.$ Because the triangle is not obtuse and $\cos B$ is negative, $\cos B = \cos (180^\circ - B) = -\frac{1}{7}.$ Then,
\[\sin B = \sqrt{1 - \left( -\frac{1}{7} \right)^2} = \frac{4 \sqrt{3}}{7}.\]Then from the Extended Law of Sines, the diameter of the circumcircle of triangle $ABC$ is
\[\frac{BC}{\sin B} = \frac{20}{\frac{4 \sqrt{3}}{7}} = \frac{35 \sqrt{3}}{2},\]so the circumradius of triangle $ABC$ is $R = \frac{BC}{2 \sin B} = \boxed{\frac{35 \sqrt{3}}{4}}.$

[asy]
unitsize(0.2 cm);

pair A, B, C;

B = (0,0);
C = (20,0);
A = 12*dir(acos(-1/7));

draw(A--B--C--cycle);
draw(rightanglemark(B,A,C,30));

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
[/asy]