Question

Two sides and an angle (SSA) of a triangle are given. Determine whether the given measurements produce one triangle, two triangles, or no triangle at all. Solve each triangle that results.  

Answer 1

In this problem, we have a triangle with:

• side a = 8,

,

• side b = 7,

,

• angle A = 30°.

Using the data of the problem, we make the following graph:

From trigonometry, we have the Law of Sines, which states that:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\text{.}$

1) Angle B

In particular, we have:

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

Replacing the values of a, b and A, we have:

$\begin{gathered} \frac{8}{\sin(30^{\circ})}=\frac{7}{\sin B}, \\ \sin B=\frac{7}{8}\cdot\sin (30^{\circ}), \\ \sin B=\frac{7}{16}\text{.} \end{gathered}$

We want to find the values of B in the interval 0 < B < 180° that satisfies the equation above. Solving the equation we get the following values:

$\begin{gathered} B_1\approx26^{\circ}, \\ B_2\approx154^{\circ}\text{.} \end{gathered}$

Now, from geometry we know that the inner angles of a triangle must sup up 180°, so the second angle B2 = 154° cannot be a possible answer, because A = 30°, and:

$A+B_2+C=30^{\circ}+154^{\circ}+C>180^{\circ}\text{.}$

So there is only one possible value for B:

$B\approx26^{\circ}\text{.}$

2) Angle C

We have the angles A = 30° and B = 26°, so angle C is:

$\begin{gathered} A+B+C=180^{\circ}, \\ C=180^{\circ}-A-B, \\ C\approx180^{\circ}-30^{\circ}-26^{\circ}, \\ C\approx124^{\circ}\text{.} \end{gathered}$

3) Side c

From the Law of Sines, we have:

$\frac{a}{\sin A}=\frac{c}{\sin C}\text{.}$

Replacing the values of a, A and C, we have:

$\begin{gathered} \frac{8}{\sin30^{\circ}}\approx\frac{c}{\sin(26^{\circ})}, \\ c\approx8\cdot\frac{\sin(26^{\circ})}{\sin(30^{\circ})}, \\ c\approx7.0 \end{gathered}$

Answers

A. There is only one possible solution to the triangle

• B ≈ 26°

,

• C ≈ 124°

,

• c ≈ 7.0