Question

Consider a triangle ABC like the one below. Suppose that =A96°, =C52°, and =b34. (The figure is not drawn to scale.) Solve the triangle.Round your answers to the nearest tenth. If there is more than one solution, use the button labeled "or".

Answer 1

We have to solve the triangle.

This means to find all the missing angles and sides.

We already know two angle measures and one side length.

We can start finding the third angle measure: we know that the sum of the angles measures has to be equal to 180°.

Then we can write:

$\begin{gathered} m\angle A+m\angle B+m\angle C=180\degree \\ 96+m\angle B+52=180 \\ m\angle B=180-96-52 \\ m\angle B=32\degree \end{gathered}$

We can now use the Law of sines to relate angles and sides.

We can write:

$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$

As we know B and b, we can use this to find the other sides as:

$\begin{gathered} \frac{\sin(A)}{a}=\frac{\sin(B)}{b} \\ a=\frac{\sin(A)}{\sin(B)}\cdot b \\ a=\frac{\sin(96\degree)}{\sin(32\degree)}\cdot34 \\ a\approx\frac{0.9945}{0.5299}\cdot34 \\ a\approx63.8 \end{gathered}$$\begin{gathered} \frac{c}{\sin(C)}=\frac{b}{\sin(B)} \\ c=\frac{\sin(C)}{\sin(B)}\cdot b \\ c=\frac{\sin(52\degree)}{\sin(32\degree)}\cdot34 \\ c\approx\frac{0.7880}{0.5299}\cdot34 \\ c\approx50.6 \end{gathered}$

Answer: B = 32°, a = 63.8, c = 50.6.