An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the radius, in inches, of the circle? Express your answer as a mixed number.
Referring to my diagram.
\(\theta+\alpha =\pi\;\;radians\\ \alpha = \pi-\theta\\ so\\ sin\alpha=sin\theta\\ cos\alpha = -cos\theta\\ \text{I am going to use the identity }cos^2\theta+\sin^2\theta=1\\\)
\(\text{Using Cosine Rule}\\ 5^2=r^2+r^2-2*r*r*cos\theta\\ 25=2r^2-2r^2cos\theta\\ \frac{25-2r^2}{-2r^2}=cos\theta\\ cos\theta=\frac{25-2r^2}{-2r^2}\\ cos^2\theta=\frac{4r^4-100r^2+625}{4r^4}\\~\\ sin\alpha=\frac{3}{r}=sin\theta\\ sin^2\theta=\frac{9}{r^2}=\frac{36r^2}{4r^4}\\~\\ Sin^2\theta+cos^2\theta=\frac{36r^2+4r^4-100r^2+625}{4r^4}=1\\ 36r^2+4r^4-100r^2+625=4r^4\\ -64r^2+625=0\\ 64r^2=625\\ r=\frac{25}{8}=3\frac{1}{8} \)
The radius of the circle is 3 and 1/8 inches.