# Area of Triangle inside a Circle in terms of angle and radius

A circle $O$ is circumscribed around a triangle $ABC$, and its radius is $r$. The angles of the triangle are $\angle CAB = a, \angle ABC = b$ and $\angle ACB = c$.

The area $\triangle ABC$ is expressed by $a, b, c$ and $r$ as:

$\Large r^2 \over\Large2$$\Bigg(\sin(x)+\sin(y)+\sin(z)\Bigg) find x, y and z: My approach: Firstly, to make it clear, I set \overline {AB} = A, \overline {BC} = B and \overline {CA} = C. \triangle ABC= \Large{Bh \over 2} where h is the height \triangle ABC = \Large{BA\sin(c) \over2} then, using the law of sine: r= \Large{A\over 2 \sin(a)} = \Large{B\over 2 \sin(b)} A = 2r\sin(a) B = 2r\sin(b) replacing on the formula of area: \triangle ABC = 2r^2\sin(a)\sin(b)\sin(c) But that doesn't help to answer the question. Is my approach correct, or else, what am I missing? • How does: "Firstly, to make it clear, I set \overline {AB} = A, \overline {BC} = B and \overline {CA} = C." make it clear? May 27 '18 at 19:13 • What are x,y,z? May 27 '18 at 19:16 • @LoveInvariants that is the question: find x,y and z: May 27 '18 at 19:17 • 2r^2sin(a)sin(b)sin(c) This is true indeed. May 27 '18 at 19:17 • Comment on notation: usually, in this kind of problem, the angles are noted A,B,C, and the sides are a,b,c, with side a opposite to the angle A (same for the other two). Also, the radius of the circumcircle is noted R, not r (which is used for the radius of the incircle). May 28 '18 at 4:41 ## 2 Answers The area formula you derived is a good one to know, but if you want something in terms of a sum of sines, use the trigonometric sum-product relations. Along the way you will also note that the angles of the rriangle sum to 180°. And be careful with signs or this won't come out pretty. Our starting point, taking S as the area: S=2r^2 \sin a \sin b \sin c Plug in \sin a \sin b =(1/2)(\cos (a-b) - \cos (a+b)) (watch signs!): S=r^2 (\cos (a-b) - \cos (a+b)) \sin c S=r^2(\cos (a-b) \sin c - \cos (a+b) \sin c) On each of these terms use \cos u \sin v =(1/2)(\sin (u+v) - \sin (u-v)) , then: S=(r^2/2)(\sin (a-b+c) - \sin (a-b-c) - \sin (a+b+c) + \sin (a+b-c)) Now for the neat part where we use the angle sum being 180°. Then, \sin (a-b+c) = \sin(180°-2b) = \sin (2b) \sin (a-b-c) = \sin(-180°+2a) = - \sin (2b) (watch signs!) \sin (a+b-c) = \sin(180°-2c) = \sin (2c) \sin (a+b+c) = \sin(180°) = 0 So we get this elegant result: S=(r^2/2)(\sin (2a) + \sin (2b) + \sin (2c)) Let X denote the center of the circle. The angles AXB, BXC and CXA are 2a, 2b, 2c. Hence$$|ABC|=|AXB|+|BXC|+|CXA|=\frac{r^2}{2}(\sin(2a)+\sin(2b)+\sin(2c))$$where$|T|$is the area of triangle$T\$. The proof works also when the triangle is obtuse - use oriented areas.

• For obtuse triangles we are required to render an angle measure greater than 180°, which does not go well with usual geometric definitions. In the long run rendering obtuse triangles as a separate case (with the offending angle measured in reverse and that triangle area explicitly subtracted) offers a "cleaner" proof. May 27 '18 at 23:24