Question

If AC=5cm, BC=12cm, and m AC= 40 degrees what is the radius of the circumscribed circle

Answer 1

Let's assume they meant C=40 degrees.  With an angle like that they're asking for approximation; we'll oblige.

The circumradius is the product of the triangle sides divided by four times the area.

Here we have remaining side given by the Law of Cosines.

$AB^2 = AC^2 + BC^2 - 2\ AC \ BC \cos C$

$AB^2 = AC^2 + BC^2 - 2 AC \ AB \cos C = 5^2 + 12^2 - 2(5)(12) \cos 40^\circ$

$AB = \sqrt{ 169 - 120 \cos 40 ^\circ} \approx 8.77921789$

The area is $\frac 1 2\ AC \ BC \sin C = \frac 1 2 (5)(12) \sin 40^\circ \approx 19.283628$



The circumradius is  $r \approx \dfrac{(5)(12)(8.77921789 )}{ 4 (19.283628) } = 6.829019329$