Chapter 6, Problem 23CR

To determine

To calculate:

Solve the triangle using law of cosine.

Answer to Problem 23CR

Triangle using law of cosine is $A=86.38°,B=28.62°,c=22.70$ .

Explanation of Solution

Given information:

  $C=65°,a=25,b=12$

Calculation:

Let’s,

  $C=65°,a=25,b=12$

For solving the triangle by using law of cosine, to determine $C$ ,

  $C^2=a^2+b^2-2ab\cos C$

Put the values in above equation,

  $\begin{array}{l}{C}^2={\left(25\right)}^2+{\left(12\right)}^2-2\left(25\right)\left(12\right)\cos \left(65°\right)\\ C^2=515.42\\ C=22.70\end{array}$

Because, $C=22.70$ , ratio known is, $\frac{\sin C}{c}$ ,

Reciprocal form of the law of sin to solve for $A$

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \sin A=a\left(\frac{\sin C}{c}\right)\end{array}$

Substitute for $a,c$ and $C$ ,

  $\begin{array}{l}\sin A=25\left(\frac{\sin \left(65°\right)}{22.7}\right)\\ \sin A=25\left(\frac{0.906330}{22.7}\right)\\ \sin A=25\left(0.03992\right)\\ \sin A=0.9981\\ A={\sin }^{-1}\left(0.9881\right)\\ A=86.38°\end{array}$

There are two angles between $0°$ and $180°$ whose $\sin$ is $0.9981$

  $\begin{array}{l}{A}_1=86.38°\\ A_2=180°-86.38°\\ A_2=93.62°\end{array}$

For $A_1=86.38°$ ,

Now, $\begin{array}{l}{B}_1=180°-\left(A_2+C\right)\\ B_2=180°-93.62°-65°\\ B_2=21.8°\end{array}$

Therefore, triangle using law of cosine is $A=86.38°,B=28.62°,c=22.70$ .