Solving a problem with the Law of Cosines gives one solution, but the Law of Sines gives two 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 


*

*angles $\angle CAO  = \angle A_2$ , $\angle
        BAO = \angle A_1$

*by the cosines law $$5^2 = 4^2+5^2-2×4×5 \cos (\angle A_1)$$
resulting that $$ \cos(\angle A_2) = 0.4 \Longrightarrow  \angle A_1
    = 66.42° $$ .

*by the cosines law $$5^2 = 6^2+5^2-2×6×5 \cos (\angle A_2)$$ resulting that $$
    \cos(\angle  A_2) = 0.6 \Longrightarrow \angle A_2 = 53.13° $$ .

*$$\angle BAC = \angle A_1 + \angle A_2 = 66.42+53.13 = 119.55$$
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 :


*

*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 $\alpha = 23.578$

*by the sine law $ \frac{6}{\sin \beta} = 10$ which gives that
$\beta_1 = 36.87 , \beta_2 = 143.13$

*triangles angles sums to $180$ so $\gamma_1 = 119.55 , \gamma_2
\Longrightarrow 13.29$

*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 ?? 
 A: Student B is correct, more or less. (The angle they're calling $\gamma_1$ corresponds to the angle they're calling $\beta_2$ and vice versa. Also, in order for them to be correct, $\gamma$ must be the angle at $A$ and $\alpha$ must be the angle at $C$; the usual convention is the other way around. But the basic idea is sound.)
When student A claims that $\angle BAC=\angle A_1+\angle A_2$, they are tacitly assuming that $\angle BAC$ contains the center of the circle. If it does not contain the center of the circle, we instead have $\angle BAC=|\angle A_1-\angle A_2|$: in this case, the larger of the angles $A_1$ and $A_2$ contains the smaller one and $\angle BAC$ is the part left over after removing it.
In that case, we would have $\angle BAC = 66.42˚-53.13˚=13.29˚$, as in student B's other solution.
A: A well-drawn figure will show that there are, in fact, two solutions:
one in which $A$ is on the shorter arc between $B$ and $C,$
and one in which $B$ is on the shorter arc between $A$ and $C.$
In the first case, $\angle BAC = \angle A_1 + \angle A_2.$
But in the second case, $\angle BAC = \angle A_1 - \angle A_2.$
The "cosine law" student neglected the solution in the case
$\angle BAC = \angle A_1 - \angle A_2.$
