In a triangle $ABC$, angle $B$ is $60$ degrees, $AB=3$, $AC=4$. Find the radius of the circumcircle of triangle $ABC$ and find $BC$.

Question

In a triangle $ABC$, angle $B$ is $60$ degrees, $AB=3$, $AC=4$. Find the radius of the circumcircle of triangle $ABC$ and find $BC$.

I put the triangle's hypotenuse on the diameter, which creates a $3-4-5$ right triangle. This means that $BC$ is $5$ and the radius is $2.5$. Am I correct?

Answer 1

Hint: Use the law of sines to find $BC$.

$$\dfrac{\sin 60}{4}=\dfrac{\sin C^\circ}{3}$$

$$\sin C^\circ=\dfrac{3{\sqrt{3}}}{8}$$

$$C\approx40.51^\circ$$

$$\angle A \approx 180-40.51-60\approx 79.49^\circ$$

Using the law of sines: $$\dfrac{4}{\sin 60}=\dfrac{BC}{\sin 79.49}$$

$$BC=\dfrac{8\sqrt{3}}{3}\cdot\sin79.49\approx4.541$$

Answer 2

Frequently students do not learn the entire version of the Law of Sines.

If $r$ is the radius of circumscribed circle of $\triangle ABC$ with sides $a,\,b,\,c$ then

$$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2r$$

So if you know a side/angle pair you can easily find the radius of the circumscribed triangle.

In this exercise you are given that $B=60^\circ$ and that $AC=b=4$.

Therefore

\begin{eqnarray} 2r&=&\frac{4}{\sin60^\circ}\\ r&=&\frac{2}{\sin60^\circ}\\ &=&\frac{4\sqrt{3}}{3} \end{eqnarray}

To find side $a=BC$ we must first find angle $C$. We know that

$$ \frac{\sin C}{3}=\frac{\sin60^\circ}{4}=\frac{\sqrt{3}}{8} $$

So either $C=\arcsin\left(\frac{3\sqrt{3}}{8}\right)\approx40.5^\circ$ or $C=180^\circ-\arcsin\left(\frac{3\sqrt{3}}{8}\right)\approx139.5^\circ$. But the second angle is too large since the triangle already has a $60^\circ$ angle. Therefore

$$C=\arcsin\left(\frac{3\sqrt{3}}{8}\right)$$

and

$$ A=180^\circ-60^\circ-\arcsin\left(\frac{3\sqrt{3}}{8}\right)=120^\circ-\arcsin\left(\frac{3\sqrt{3}}{8}\right) $$

So \begin{eqnarray} \sin A&=&\sin120^\circ\cos\left(\arcsin\left(\frac{3\sqrt{3}}{8}\right)\right)-\cos120^\circ\cdot\left(\frac{3\sqrt{3}}{8}\right)\\ &=&\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{37}}{8}-\left(-\frac{1}{2}\right)\cdot\left(\frac{3\sqrt{3}}{8}\right)\\ &=&\frac{\sqrt{3}(\sqrt{37}+3)}{16} \end{eqnarray}

Then from the Law of Sines we get

$$a=BC=2r\sin A=2\cdot\frac{4\sqrt{3}}{3}\cdot\frac{\sqrt{3}(\sqrt{37}+3)}{16}=\frac{\sqrt{37}+3}{2}$$

Answer 3

By the sine rule

\begin{align} R&=\frac{|AC|}{2\,\sin\angle ABC} =\frac{4}{2\cdot\tfrac{\sqrt3}2} =\frac{4\sqrt3}{3} \approx 2.3 . \end{align}

By the cosine rule

\begin{align} |AC|^2&=|AB|^2+|BC|^2-2\,|AB|\cdot|BC|\cdot\cos\angle ABC \end{align} we have a quadratic equation in terms of $|BC|$,

\begin{align} |BC|^2-3\,|BC|-7&=0 , \end{align}
which has one suitable (positive) solution,

\begin{align} |BC|&=\tfrac32+\tfrac{\sqrt{37}}2 \approx 4.54 . \end{align}