# Area of a Triangle Inside a Circle?

I have a circle of radius $r$ and a triangle inside that circle. Specifically, if you have the triangle $\triangle ABC$ inside a circle with only the one side $AB$ and an angle $\angle \text{B}$ opposite to the other side $AC$ known along with circle radius $r$, could you determine its area? • can you post a Picture please? Dec 7, 2015 at 17:49
• Can we make the following assumptions? 1. $\triangle ABC$ is inscribed in the circle; that is, $A, B, C$ all lie on the circumference. 2. We know the length $AB$ and the angle measure $m\angle ABC$. Dec 7, 2015 at 17:52
• Those data are not enough to find the area of the triangle. If you have a certain cord of the circle, then the opposite angle will be the same no matter where on the circle the third corner of the triangle is -- namely $\sin^{-1}\frac{|AB|}{2r}$ Dec 7, 2015 at 17:54
• If those assumptions are valid, the area of the triangle can be determined. What are you able to use, mathematically? (The tools you have at your disposal determine what kind of solution can be presented.) Dec 7, 2015 at 17:55
• @Brian: Ah, I see. That seems to be a maximally confusing way of saying it rather than "one side and one of its adjacent angles known". Dec 7, 2015 at 17:57

The angle at $C$ must be $\arcsin\frac{|AB|}{2r}$, namely half of the arc spanned by $AB$. This gives you all the angles in the triangle.

The law of sines then gives you the sides.

And then there are many ways to get the area, such as Heron's formula, or the sine of one angle times half the product of the adjacent sides.

• Of course there are two possibilities for the angle (supplementary), nevertheless $\sin$ of the angle is known. The other angles can vary. Dec 7, 2015 at 18:05
• Is there a way to solve this without calculating the arcsin? Like if you were forbidden to use a calculator? Dec 7, 2015 at 18:07
• Do we know $|AB|$ at the beginning!? I think it should be computed from the inputs of the problem, i.e., $\theta$, $a$, $r$ as showed in the picture. Dec 7, 2015 at 18:12
• I found a way to solve it without using inverse trigonometric formulas. But I am pretty sure you won't be able to get away with out using normal trigonometric formulas. Dec 7, 2015 at 18:14
• @H.R. The question did say "one side ... known". Dec 7, 2015 at 18:14

Let $O$ be the center of the circle.

Let $\alpha=\angle ABO$ and $\beta=\angle OBC$

Then $\sin(\angle B)=\sin(\alpha)\cdot \cos(\beta)+\sin(\beta)\cdot \cos(\alpha)$.

Of course $\cos(\alpha)=\frac{|AB|}{2r}$ and $\sin(\alpha)=\sqrt{1-\cos^2(\alpha)}$.

Therefore if $x=\cos(\beta)$ we get

$\sin(\angle B)=\sqrt{1-(\frac{|AB|}{2r})^2}\cdot x + \frac{|AB|}{2r}\sqrt{1-x^2}$.

So we solve the equation for $x$ to get $\cos(\beta)$ (don't get scared, if you take the first summand to the left side and square it is just a quadratic polynomial in $x$).

And once we find $\cos(\beta)$ we have $|BC|=2r\cdot \cos(\beta)$

Finally, the area is just $\sin(\angle B)\cdot\frac{|BC||BA|}{2}$

• could i have used the sine rule to get the $AC$ side? Then by the means of the cosin rule get the other side. Then calculating the area would be easy Dec 8, 2015 at 8:10
• you can surely use the sine rule to get $AC$. But getting side $|BC|$ using the cosine law is not possible. Dec 8, 2015 at 14:02