Let $ABC$ be triangle inscribe in circle with radius $5$ and center $O$, such that $AB = 4$ and $AC =6$ and the question is to find $BC = ?$

* **Student A method** 

	1. angles $\angle CAO = \angle A\_2$ , $\angle
	 BAO = \angle A\_1$
	2. by the cosines law $$5^2 = 4^2+5^2-2×4×5 \cos (\angle A\_1)$$
	resulting that $$ \cos(\angle A\_2) = 0.4 \Longrightarrow \angle A\_1
	 = 66.42° $$ .
	3. by the cosines law $$5^2 = 6^2+5^2-2×6×5 \cos (\angle A\_2)$$ resulting that $$
	 \cos(\angle A\_2) = 0.6 \Longrightarrow \angle A\_2 = 53.13° $$ .
	4. $$\angle BAC = \angle A\_1 + \angle A\_2 = 66.42+53.13 = 119.55$$
	by the cosines law $$BC^2 = 4^2+6^2-2×4×6 \cos {119.55}$$ resulting that $BC^2 = 75.67$ and so $BC = 8.7$

.

* **Student B method :**

	1. by the sine law $\frac{4}{\sin \alpha} = 2r = 2×5 = 10$ so $ \sin
	\alpha = 0.4$ so $\alpha = 23.578$ or $\alpha = 156.42$ which is
	invalid because the opposite angle of smaller side is smaller and
	since $4<6$ then $\alpha = 23.578$
	2. by the sine law $ \frac{6}{\sin \beta} = 10$ which gives that
	$\beta\_1 = 36.87 , \beta\_2 = 143.13$
	3. triangles angles sums to $180$ so $\gamma\_1 = 119.55 , \gamma\_2
	\Longrightarrow 13.29$
	4. by the sine law $\frac{BC}{\sin\gamma} =10 $ so $BC = 10 \sin
	\gamma\_1 = 2.29$ or $BC = 10 \sin \gamma\_2 = 8.7$

Which one of the methods are the correct one ??