# Let $ABC$ be a triangle with $AB=AC=6$. If the circumradius of the triangle is $5$ then $BC$ equals:___

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

• $\angle BAC=106^\circ$ — Redraw the diagram with an obtuse triangle and you'll figure it out.
– dxiv
Jul 3 at 8:00
• i didn't get it Jul 3 at 8:32
• you took $tan\theta=4/3$ and $\theta = 53$ which a approximate value just to solve physics problem. Jul 3 at 9:28
• Do you know about formula $R=\frac{abc}{4 \cdot Area of Triangle}$ ? Jul 3 at 9:29
• Is length of BC $9.6$ Jul 3 at 9:30

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, $$BE = \dfrac{BC}{2} = AB \cos \angle B = \dfrac{24}{5}$$

$$BC = \dfrac{48}{5}$$

Usual notational conventions for the sides of a triangle are used

Suppose $$\delta$$ denotes the area of $$ABC$$ and $$R$$ the circumradius Also let's define $$\angle BAC=\xi$$

Then we've $$\frac{abc}{4R}=\frac{bc \sin(\xi)}{2}=\delta$$

From this we get using the Law of sines

$$\frac{a}{\sin(\xi)}=10=\frac{b}{\sin\left(\frac{\pi}{2}-\frac{\xi}{2} \right)}=\frac{b}{\cos \left(\frac{\xi}{2} \right)}$$

Thus we get $$\cos(\xi)=\left(2\cos\left(\frac{\xi}{2} \right)\right)^{2}-1=\frac{-7}{25}$$

Using the law of cosines we get $$\cos(\xi)=\frac{b^2 +c^2-a^2}{2bc}$$ Solving this we get $$a=\frac{48}{5}$$