In triangle ABC, (cos B + cos A cos C)/(sin B cos C) = a sqrt(3)/b, with a = 3 and c = 2 sqrt(3). Find sin A.

Step-by-step solution

1. 1

   Replace $\cos B$ using the angle sum of the triangle. In any triangle $A+B+C=\pi$, so $B=\pi-(A+C)$ and

   $\cos B=\cos\!\left(\pi-(A+C)\right)=-\cos(A+C)=-\left(\cos A\cos C-\sin A\sin C\right).$

   This substitution is the standard first move whenever a triangle identity mixes all three angles.

2. 2

   Collapse the numerator. Adding $\cos A\cos C$ makes the two $\cos A\cos C$ terms cancel:

   $\cos B+\cos A\cos C=-\cos A\cos C+\sin A\sin C+\cos A\cos C=\sin A\sin C.$

   So the left-hand side becomes

   $\frac{\sin A\sin C}{\sin B\cos C}=\frac{\sin A}{\sin B}\cdot\tan C.$

3. 3

   Use the law of sines to match the right-hand side. Since $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$, we have $\dfrac{\sin A}{\sin B}=\dfrac{a}{b}$, so the condition reads

   $\frac{a}{b}\tan C=\frac{a\sqrt3}{b}.$

   Side lengths are positive, so the factor $\dfrac{a}{b}$ cancels and

   $\tan C=\sqrt3\ \Longrightarrow\ C=\frac{\pi}{3}=60^{\circ},$

   the only solution with $0<C<\pi$ and $\cos C\ne 0$. Notice the given identity determines $C$ alone — the side data was never needed for this part.

4. 4

   Apply the law of sines with the known pair $(c,C)$.

   $\frac{a}{\sin A}=\frac{c}{\sin C}\ \Longrightarrow\ \sin A=\frac{a\sin C}{c}=\frac{3\cdot\frac{\sqrt3}{2}}{2\sqrt3}=\frac{3\sqrt3}{4\sqrt3}=\frac{3}{4}.$

5. 5

   Check that the triangle really exists and is unique. Since $a=3<c=2\sqrt3\approx 3.464$, angle $A$ is smaller than $C=60^{\circ}$, so $A=\arcsin\tfrac34\approx 48.59^{\circ}$ and the obtuse alternative $131.41^{\circ}$ is impossible (it would leave no room for $C$). Then $B\approx 71.41^{\circ}$ and $b=\dfrac{c\sin B}{\sin C}\approx 3.7913=\dfrac{3+\sqrt{21}}{2}$; substituting these values back into the original expression gives $1.37055$ on both sides.