Circumradius from Law of Cosines In , given , , and , find the circumradius of .

Trigonometry Solution – Problem 23: find the circumradius of

Circumradius from Law of Cosines

In $\triangle ABC$ , given $c = 1$ , $b = 2$ , and $A = 60°$ , find the circumradius $R$ of $\triangle ABC$ .

Using the Law of Cosines to find $a$ :

$a^2 = b^2 + c^2 - 2bc\cos A$

$a^2 = 4 + 1 - 2 \cdot 2 \cdot 1 \cdot \frac{1}{2} = 3$

$a = \sqrt{3}$

Using the Law of Sines for circumradius:

$2R = \frac{a}{\sin A} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2$

$R = 1$

$1$

Choose the correct theorem (2 pts): Law of Sines / Law of Cosines / area formula / circumradius relation as appropriate.
Set up equations correctly (2 pts): substitute given data and write a solvable system.
Solve and (if needed) reject extraneous cases (2 pts): handle SSA ambiguity or inequality constraints if present.
Final answer (1 pt): provide the requested length/angle/area in the required form.