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

Step-by-step solution

1. 1

   Use the angle sum to eliminate B from the numerator. In any triangle $A+B+C=\pi$, so $B=\pi-(A+C)$ and

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

   This substitution is the entire trick — it is what makes the awkward numerator collapse.

2. 2

   Simplify the numerator.

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

   So the left-hand side becomes

   $\frac{\sin A\sin C}{\sin B\cos C}.$

3. 3

   Convert the right-hand side to sines with the law of sines. Since $\dfrac{a}{b}=\dfrac{\sin A}{\sin B}$,

   $\frac{a\sqrt{3}}{b}=\frac{\sqrt{3}\sin A}{\sin B}.$

4. 4

   Cancel and solve for C. Equating the two sides and cancelling the common factor $\dfrac{\sin A}{\sin B}$ (non-zero in a triangle):

   $\frac{\sin C}{\cos C}=\sqrt{3}\ \Longrightarrow\ \tan C=\sqrt{3}\ \Longrightarrow\ C=\frac{\pi}{3},$

   since $C$ lies in $(0,\pi)$ and $\cos C\neq 0$ for the original expression to be defined.

5. 5

   Apply the law of sines with the given sides. With $a=3$, $c=2\sqrt3$ and $\sin C=\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2}$:

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

6. 6

   Sanity-check the triangle. Since $a=3<c=2\sqrt3\approx 3.464$, angle $A$ must be smaller than $C=60^\circ$; indeed $\arcsin\tfrac34\approx 48.6^\circ<60^\circ$, so the acute value is the right one and no ambiguous second solution survives.