Chapter 5.7, Problem 8CFU

To determine

To find: All solutions for triangle $A={38}^{\circ },b=10,a=8$ . If no solutions exist, write none . Round to the nearest tenth

Answer to Problem 8CFU

  $\begin{array}{l}\angle B={50.3}^{\circ },\angle C={91.7}^{\circ },c=13\text{ this is for the first triangle}\\ \angle B={129.7}^{\circ },\angle C={12.3}^{\circ },c=2.8\text{ for the second triangle}\end{array}$

Explanation of Solution

Given information: $A={38}^{\circ },b=10,a=8$

Calculation:

This is our triangle, since A is acute we compare a to b . a < b. We need to find $b\sin A$ to test for the three possibilities.

  $10\sin {38}^{\circ }=6.2\text{ which is less than }a\text{ so}$ ,

There are two triangles to be formed. Let us go with the one drawn first.

  $\begin{array}{l}\frac{8}{\sin {38}^{\circ }}=\frac{10}{\sin B}\\ \sin B=\frac{10\times \sin {38}^{\circ }}{8}=0.7696\\ {\sin }^{-1}0.7696=\angle B={50.3}^{\circ }\\ \angle C={180}^{\circ }-\left({38}^{\circ }+{50.3}^{\circ }\right)={91.7}^{\circ }\end{array}$

Now use the law of sines again to find side c

  $\begin{array}{l}\frac{8}{\sin {38}^{\circ }}=\frac{c}{\sin {91.7}^{\circ }}\\ c=\frac{8\times \sin {91.7}^{\circ }}{\sin {38}^{\circ }}=13\end{array}$

Here is the other possible triangle; angle B in this triangle will be supplementary to angle B in the first triangle or the new angle $\angle B={180}^{\circ }-{50.3}^{\circ }={129.7}^{\circ }$

Now use the law of sines

  $\begin{array}{l}\angle C={180}^{\circ }-\left({129.7}^{\circ }+{38}^{\circ }\right)={12.3}^{\circ }\\ \frac{8}{\sin {38}^{\circ }}=\frac{c}{\sin {12.3}^{\circ }}\\ c=\frac{8\times \sin {12.3}^{\circ }}{\sin {38}^{\circ }}=2.8\end{array}$

Thus,

  $\begin{array}{l}\angle B={50.3}^{\circ },\angle C={91.7}^{\circ },c=13\text{ this is for the first triangle}\\ \angle B={129.7}^{\circ },\angle C={12.3}^{\circ },c=2.8\text{ for the second triangle}\end{array}$