Chapter 5.7, Problem 2CFU

To determine

To draw: Two triangles where $A={30}^{\circ },a=6\text{ and }b=10$ . Calculate and label the degree measure of each angle rounded to the nearest tenth.

Answer to Problem 2CFU

Two triangles

Explanation of Solution

Given information: $A={30}^{\circ },a=6\text{ and }b=10$

Calculation:

Here are our two triangles. One has side a going away and one has side a coming back to us. This is because $b>a>b\sin a$ (the numbers and math gives us 10 > 6 > 5 (found from $10\sin {30}^{\circ }$ ))

For the left triangle, this will give us angle B and allow us to find angle C

  $\begin{array}{l}\frac{6}{\sin {30}^{\circ }}=\frac{10}{\sin B}\\ \sin B=\frac{10\sin {30}^{\circ }}{6}=\frac{5}{6}\end{array}$

We need the arc sine

Now the second triangle, if B = 56.4 in the first triangle the B value in the other triangle will be its supplement or ${180}^{\circ }-{56.4}^{\circ }={123.6}^{\circ }$ and angle C will be ${180}^{\circ }-\left({123.6}^{\circ }+{30}^{\circ }\right)={26.4}^{\circ }$

  $\begin{array}{l}{\sin }^{-1}\left(\frac{5}{6}\right)=\angle B={56.4}^{\circ },\\ \angle C={180}^{\circ }-\left({30}^{\circ }+{56.4}^{\circ }\right)={93.6}^{\circ }\end{array}$

Here are the triangles redrawn with all angles labeled.