algebra-precalculusgeometrytrigonometry
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.
I'll describe my methods:
1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.
2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).
Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.
3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$
Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.
At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $ ∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.
4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.
$GE=1/2*CE (opposite 30), ACE isosceles (angles 70,70),draw perpendicular to CE, there are 2 congruent right triangles , angle 20, common hypotenuse. So, GE=EI.
Best Answer
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 $$\frac{1}{2}\cdot\sqrt 6\ (\sqrt 2+\sqrt 6)=3+\sqrt 3.$$