Step 1. We know the measure of all of the angles in the triangle and the measure of side b:
Required: Find the measure of sides a and c to the nearest tenth.
Step 2. To find a and c we use the law of sines.
A, B, and C are the angles of the triangle and a, b, and c are the sides. The law of sines is:
[tex]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}[/tex]
In our case, substituting the known values:
[tex]\frac{a}{sin60}=\frac{53}{sin42}=\frac{c}{sin78}[/tex]
Step 3. Using the first equality we can find the value of a:
[tex]\frac{a}{s\imaginaryI n60}=\frac{53}{s\imaginaryI n42}[/tex]
Solving for a:
[tex]a=\frac{53}{s\imaginaryI n42}\times sin60[/tex]
Solving the operations:
[tex]\begin{gathered} a=\frac{53}{(0.669130606)}(0.866025403) \\ \downarrow \\ \boxed{a=68.6} \end{gathered}[/tex]
The result is already rounded to the nearest tenth.
Step 4. Using the second equality from step 2
[tex]\frac{53}{sin42}=\frac{c}{sin78}[/tex]
We will be able to find the value of c.
Solving for c:
[tex]sin78\times\frac{53}{sin42}=c[/tex]
Solving the operations:
[tex]\begin{gathered} (0.9781476)\frac{53}{(0.669130606)}=c \\ \downarrow \\ \boxed{77.5=c} \end{gathered}[/tex]
The result is already rounded to the nearest tenth.
Answer:
[tex]\begin{gathered} \boxed{a=68.6} \\ \boxed{c=77.5} \end{gathered}[/tex]
