Chapter 6, Problem 24CR

To determine

To calculate:

Solve the triangle using law of cosine.

Answer to Problem 24CR

Triangle using law of cosine is $A=3.57°,B=128.43°,b=214.40$ .

Explanation of Solution

Given information:

  $B=48°,a=8,c=18$

Calculation:

Let’s,

  $B=48°,a=8,c=18$

Grah for $B=48°,a=8,c=18$ ,

  

For solving the triangle by using law of cosine,

  $b^2=a^2+c^2-2ac\cos B$

Put the values in above equation,

  $\begin{array}{l}{b}^2={\left(18\right)}^2+{\left(18\right)}^2-2\left(18\right)\left(18\right)\cos \left(48°\right)\\ b^2=648-433.59\\ b=214.40\end{array}$

Because, $b=214.40$ , ratio known is, $\frac{\sin B}{b}$ ,

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

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

Substitute for $a,b,c$ and $B$ ,

  $\begin{array}{l}\sin A=18\left(\frac{\sin \left(48°\right)}{214.40}\right)\\ \sin A=18\left(\frac{0.7431}{214.40}\right)\\ \sin A=0.062390\\ A={\sin }^{-1}\left(0.062390\right)\\ A=3.5770°\end{array}$

There are two angles between $0°$ and $180°$ whose $\sin$ is $0.062390°$

  $\begin{array}{l}{A}_1=3.57°\\ A_2=180°-3.57°\\ A_2=176.42°\end{array}$

For $A_1=3.57°$ ,

Now,

  $\begin{array}{l}{C}_1=180°-\left(A_2+C\right)\\ C_1=180°-3.57°-48°\\ C_1=128.43°\end{array}$

For $A_2=176.42°$ ,

  $\begin{array}{l}{C}_2=180°-\left(A_2+C\right)\\ C_2=180°-176.42°-48°\\ C_2=-44.42°\end{array}$

Because side $a$ is the longest side of the triangle $A$ must be largest angle of the triangle,

Therefore, triangle using law of cosine is $A=3.57°,B=128.43°,b=214.40$ .