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

  • 2
    $\begingroup$ $\angle BAC=106^\circ$ — Redraw the diagram with an obtuse triangle and you'll figure it out. $\endgroup$
    – dxiv
    Jul 3 at 8:00
  • $\begingroup$ i didn't get it $\endgroup$ Jul 3 at 8:32
  • $\begingroup$ you took $tan\theta=4/3$ and $\theta = 53$ which a approximate value just to solve physics problem. $\endgroup$
    – mathophile
    Jul 3 at 9:28
  • $\begingroup$ Do you know about formula $R=\frac{abc}{4 \cdot Area of Triangle}$ ? $\endgroup$
    – mathophile
    Jul 3 at 9:29
  • $\begingroup$ Is length of BC $9.6$ $\endgroup$
    – mathophile
    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 $a=\frac{48}{5}$


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