Given that in a triangle ABC, AB = 2cm, AC = 3cm, $\displaystyle \angle A =60^\circ $ Find the radius of the circumscibed circle of ABC.

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- Aug 4th 2012, 03:06 AMshiny718Find the radius of the circumscribed circle
Given that in a triangle ABC, AB = 2cm, AC = 3cm, $\displaystyle \angle A =60^\circ $ Find the radius of the circumscibed circle of ABC.

- Aug 4th 2012, 03:19 AMProve ItRe: Find the radius of the circumscribed circle
We know that the three vertices of the triangle lie on a circle. If we say that point A is the origin (0, 0), then we could say that since the length AC is 3 cm, the co-ordinate of C is (3, 0).

We know that point B lies 2cm from point A making an angle of $\displaystyle \displaystyle \begin{align*} 60^{\circ} \end{align*}$.

The vertical position of point B can be found by

$\displaystyle \displaystyle \begin{align*} \sin{60^{\circ}} &= \frac{y}{2} \\ \frac{\sqrt{3}}{2} &= \frac{y}{2} \\ y &= \sqrt{3} \end{align*}$

and the horizontal position of point B can be found by

$\displaystyle \displaystyle \begin{align*} \cos{60^{\circ}} &= \frac{x}{2} \\ \frac{1}{2} &= \frac{x}{2} \\ x &= 1 \end{align*}$

So point B is at $\displaystyle \displaystyle \begin{align*} \left(1, \sqrt{3}\right) \end{align*}$.

Now that you have three points that lie on your circle, you can substitute them into the general equation for a circle $\displaystyle \displaystyle \begin{align*} (x - h)^2 + (y - k)^2 = r^2 \end{align*}$ and solve them simultaneously for $\displaystyle \displaystyle \begin{align*} h, k, r \end{align*}$ (you are trying to find r). - Aug 4th 2012, 09:21 AMrichard1234Re: Find the radius of the circumscribed circle
You can find BC either by using law of cosines or by constructing an altitude from point B and using 30-60-90 triangles. Either way, you should get $\displaystyle BC = \sqrt{7}$.

There is a formula for the area of a triangle, $\displaystyle [ABC] = \frac{AB*BC*CA}{4R}$, where R is the circumradius. We know AB, BC, CA, and just need to find the area. This is easy, since

$\displaystyle [ABC] = \frac{1}{2}AB*BC*\sin{A} = \frac{1}{2}(2)(3)(\frac{\sqrt{3}}{2}) = \frac{3 \sqrt{3}}{2}$

Hence, $\displaystyle \frac{3 \sqrt{3}}{2} = \frac{2*3*\sqrt{7}}{4R}$

$\displaystyle 12R \sqrt{3} = 12 \sqrt{7}$

$\displaystyle R = \frac{\sqrt{7}}{\sqrt{3}} = \frac{\sqrt{21}}{3}$.