This is the question.
Let $ABC$ be a triangle with $AB=AC=6$. If the circumradius of the triangle is $5$ then $BC$ equals:___
My approach:
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
Best Answer
As you found $\angle BAC$ is $106^0$ and as it is an obtuse angled triangle, the circumcenter of $\triangle ABC$ is outside the triangle as mentioned in the comments. If $E$ is the midpoint of $BC$, circumcenter is on line going through $AE$, outside of triangle and below $BC$. So yes $\angle BCA = 37^0$ and $\angle OCA = 53^0$ are both possible.
But you can make your working much simpler by applying extended sine rule.
$\dfrac{AC}{\sin \angle B} = 2 R \implies \sin \angle B = \dfrac{6}{10} = \dfrac{3}{5}$
So, $\cos \angle B = \dfrac{4}{5}$
If $E$ is the midpoint of $BC$, $\triangle AEB$ is right angled triangle.
So, $BE = \dfrac{BC}{2} = AB \cos \angle B = \dfrac{24}{5}$
$BC = \dfrac{48}{5}$