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Points $A$, $B$, and $C$ are on the circumference of a circle with radius $2$ such that $\angle BAC = 45^\circ$ and $\angle ACB = 60^\circ$. Find the area of $\triangle ABC$.

I've drawn a circle with radius $2$ and drew the triangle. I tried to construct a few more triangles inside the triangle to solve it but it didn't seem to help.

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  • $\begingroup$ do you know the area formula for a triangle, if given length of three sides. $\endgroup$ – Yimin Jul 6 '15 at 19:50
  • $\begingroup$ Yes I do. But what are the side lengths? Thats what I am trying to find out. $\endgroup$ – SAM Jul 6 '15 at 19:54
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By the law of sines, we have $$\frac{\overline{BC}}{\sin\angle{BAC}}=2\cdot 2\Rightarrow \overline{BC}=2\sqrt 2.$$ Then, let $D$ be a point on the side $CA$ such that $BD$ is perpendicular to $CA$.

Here, note that $\triangle{ABD}$ is a triangle with $45^\circ,45^\circ,90^\circ$ and that $\triangle{BCD}$ is a triangle with $30^\circ,60^\circ,90^\circ$.

Now since $\overline{CD}=\sqrt 2,\overline{BD}=\overline{AD}=\sqrt 6,$ the area of $\triangle{ABC}$ is $$\frac{1}{2}\cdot\sqrt 6\ (\sqrt 2+\sqrt 6)=3+\sqrt 3.$$

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Consider the figure below:

circumscribed_triangle

We can find side lengths $a$ and $c$ using the Law of Sines:

$$\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R$$

where $R = 2$ is the radius of the circumscribed circle.

Since $\alpha = 45^\circ$, \begin{align*} a & = 2R\sin\alpha\\ & = 2 \cdot 2 \cdot \sin(45^\circ)\\ & = 4 \cdot \frac{\sqrt{2}}{2}\\ & = 2\sqrt{2} \end{align*} Since $\gamma = 60^\circ$, \begin{align*} c & = 2R\sin\gamma\\ & = 2 \cdot 2 \cdot \sin(60^\circ)\\ & = 4 \cdot \frac{\sqrt{3}}{2}\\ & = 2\sqrt{3} \end{align*} The area of the triangle is $$A = \frac{1}{2}ac\sin\beta$$ since $c\sin\beta$ is the length of the altitude to side $\overline{BC}$, which has length $a$.

By the Angle Sum Theorem for Triangles, \begin{align*} \beta & = 180^\circ - 60^\circ - 45^\circ\\ & = 75^\circ \end{align*}

To find the exact value of $\sin(75^\circ)$, we use the Sum of Angles Formula $$\sin(\theta + \varphi) = \sin\theta\cos\varphi + \cos\theta\sin\varphi$$ with $\theta = 30^\circ$ and $\varphi = 45^\circ$, which yields \begin{align*} \sin(75^\circ) & = \sin(30^\circ)\cos(45^\circ) + \cos(30^\circ)\sin(45^\circ)\\ & = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\\ & = \frac{\sqrt{2} + \sqrt{6}}{4} \end{align*} Hence, the area of triangle $ABC$ is \begin{align*} A & = \frac{1}{2}ac\sin\beta\\ & = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2\sqrt{3} \cdot \frac{\sqrt{2} + \sqrt{6}}{4}\\ & = \frac{1}{2} \cdot \sqrt{6}(\sqrt{2} + \sqrt{6})\\ & = \frac{1}{2}(\sqrt{12} + 6)\\ & = \frac{1}{2}(2\sqrt{3} + 6)\\ & = \sqrt{3} + 3 \end{align*} square units.

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If $O$ is the circumcenter of $ABC$, we have:

$$ \Delta = [OAB]+[OAC]+[OBC] = \frac{R^2}{2}\left(\sin(2A)+\sin(2B)+\sin(2C)\right).\tag{1} $$ In our case, we have $A=45^\circ,C=60^\circ$, hence $B=75^\circ$ and: $$ \Delta = 2\left(\sin 90^\circ+\sin 120^\circ + \sin 150^\circ\right)=2\left(1+\frac{\sqrt{3}}{2}+\frac{1}{2}\right)=\color{red}{3+\sqrt{3}}.\tag{2} $$

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Let the Center of the Circle be O. Since the internal triangles you have drawn are isosceles, we can determine each of the internal angles.We have the following system of equations: let $\angle{OAB} = \angle{OBA} = x$

$\angle{OAC} + \angle{OCA} = y$

$\angle{OCB} + \angle{OBC} = z$

It follows that:

$x+z = 75$

$x+y = 45$

$y+z = 60$

Solving for each variable gives:

$x = 30$

$y= 15$

$z=45$

Therefore, we can determine the central angles to be: $\angle{AOC} = 150$

$\angle{COB} = 90$

$\angle{BOA} = 120$

The area of the triangle is: $\frac{2*2 \sin{\angle{OAC}}}{2} + \frac{2*2 \sin{\angle{COB}}}{2} + \frac{2*2 \sin{\angle{BOA}}}{2}$ =

$\frac{2*2 \sin{150}}{2} + \frac{2*2 \sin{90}}{2} + \frac{2*2 \sin{120}}{2} = $

$1 + 2 + \sqrt{3}$ =

$3 + \sqrt{3}$

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A circular arc with central angle $\theta$ subtends the angle $\theta/2$ from any other point on the circle (not a point on the arc). Conversely, if two points on a circle subtend an angle $\theta$ from another point on the circle, they subtend a central angle of $2\theta$ along the same arc intercepted by the original angle.

Let $O$ be the center of your circle. Since $A$, $B$, and $C$ are all points on the circle, we find that $\angle BOC = 90^\circ$ (because $\angle BAC = 45^\circ$) and $\angle AOB = 120^\circ$ (because $\angle ACB = 60^\circ$).

From these facts you can easily find $\angle AOC$.

If you then draw the triangles $\triangle AOB$, $\triangle AOC$, and $\triangle BOC$, you will see that they partition the triangle $\triangle ABC$, so now all you need to do is add up their areas. They are all isoceles triangles with two sides equal to $2$.

The rest of the necessary calculations are in Jack D'Aurizio's answer, where $R=2$ is the length of the equal sides of the isoceles triangles.

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