Question

11) Using the law of sines, determine whether the given information results in one triangle, two triangle or no triangle at all. Solve any triangle that results. a=26b = 29Angle A =58 degree

Answer 1

To determine the number of solutions, let us first solve for h.

$\begin{gathered} h=a\sin B \\ h=26(\sin29) \\ h=12.6 \end{gathered}$

We will also need to find b using the sine law.

$\begin{gathered} \frac{a}{sinA}=\frac{29}{sinB} \\ \\ \frac{26}{sin58}=\frac{29}{sinB} \\ \\ \sin B=\frac{29(\sin58)}{26} \\ \sin B=0.946 \\ \sin^{-1}(\sin B)=\sin^{-1}0.946 \\ B=71.07 \end{gathered}$

Here we see that a = 26, b = 29, and h = 12.6.

When h < a < b, then are 2 solutions or 2 triangles.

Next, we solve for the 2 solutions.

For the first solution, we already calculated B to be 71.07 degrees. We know that A = 58 degrees. So to find C, we simply subtract their sum from 180.

$180-(58+71.07)=50.93$

Again, using the sine law, we can solve for c.

$\begin{gathered} \frac{a}{sinA}=\frac{c}{sinC} \\ \\ \frac{26}{sin58}=\frac{c}{sin50.93} \\ \\ c=\frac{26(sin50.93)}{sin58} \\ \\ c=23.80 \end{gathered}$

The other solution is when B = 180 - 71.07 = 108.93. We do this because there are 2 angle measurements that have the same sine (in this case, it's 0.946).

Again, the other possible value of B is 180 - 71.07 or 108.93. That will give us C = 180 - (53 + 108.93) = 18.07 degrees.

Again, we use the sine law to solve for the corresponding c.

$\begin{gathered} \frac{26}{\sin58}=\frac{c}{\sin18.07} \\ \\ c=\frac{26(\sin18.07)}{\sin58} \\ \\ c=9.51 \end{gathered}$

So the answers are:

1. There are 2 possible triangles.

2. Triangle 1: B = 71.07 degrees, C = 50.93 degrees, and c = 23.80.

3. Triangle 2: B = 108.93 degrees, C = 18.07 degrees, and c = 9.51