This is the question.
Let $ABC$ be a triangle with $AB=AC=6$. If the circumradius of the triangle is $5$ then $BC$ equals:___
This is the diagram I drew (it's a bit messy, sorry for that).
$(1)$ Since O is the Circumcentre; $OA=OB=OC=5 \ cm$
As given in the question.
$(2)$ I wanted to find $\angle OCD$ in order to find the remaining angles and get the length of $BC$ in form of a trigonometric ratio.
$(3)$ So, I used Heron's formula to calculate the area of $\triangle AOC$, which I got as $12 \ cm^2$
$(4)$ I then calculated the height $OD$ which I got as $4 \ cm$
$(5)$ By trigonometric relations, $\angle OCD$ comes out to be $53$ degrees.
Since, $AO=OC$, $\angle OAC$ must be $53$ degrees too.
And even $\angle BAO= 53$ degrees.
Therefore, $\angle BAC = 106$ degrees.
Since, $\Delta ABC$ is also Isosceles, we have:
$\angle ABC = \angle ACB = x $ (let)
$\therefore 2x+106 = 180 \implies x = 37$
But we have $\angle ABO = \angle OBC = 53$ degrees!
And it is impossible for $\angle ABO$ to be greater than $\angle ABC$
Could someone tell where I am making a mistake? Thanks