In a triangle, the side of length 13 is opposite a 32 degree angle, the side of length 23 is opposite angle A, and side x is opposite the third angle. Assuming A is acute, find angle A and the length of x.

Step-by-step solution

1. 1

   Match each side with the angle across from it. The law of sines pairs a side with its opposite angle:

   $\frac{\sin A}{23}=\frac{\sin 32^\circ}{13}=\frac{\sin C}{x}.$

   Getting a side paired with the wrong angle is the main failure mode here — always check 'opposite', not 'adjacent'.

2. 2

   Solve the first pair for sin A.

   $\sin A=\frac{23\sin 32^\circ}{13}=\frac{23(0.529919)}{13}=\frac{12.18815}{13}=0.937549.$

   That the value came out below $1$ is itself information: had it exceeded $1$, no such triangle would exist.

3. 3

   Take the inverse sine — and confront the ambiguity. The calculator returns

   $A=\arcsin(0.937549)=69.64^\circ,$

   but $\sin(180^\circ-69.64^\circ)=\sin(110.36^\circ)$ is the same value, so an obtuse triangle is also arithmetically possible. This is the SSA ambiguous case. The problem states that an angle appearing acute may be assumed acute, so take

   $A\approx 69.6^\circ.$

4. 4

   Find the third angle.

   $C=180^\circ-32^\circ-69.64^\circ=78.36^\circ.$

5. 5

   Use the law of sines again for x.

   $x=\frac{13\sin C}{\sin 32^\circ}=\frac{13(0.979424)}{0.529919}=\frac{12.73251}{0.529919}=24.03.$

6. 6

   Check the ordering of sides and angles. The largest angle is $C=78.4^\circ$ and the largest side is $x=24.03$; the smallest angle is $32^\circ$ and the smallest side is $13$. Sides and angles are in the same order, as they must be. (For the record, the rejected obtuse branch would have given $A=110.4^\circ$ and $x=14.98$.)