Given:
Angle B = 103 degrees
Angle C = 28 degrees
Side b = 52 units
Find: measures of Angle A, side a, and side c.
Solution:
First, let's solve for the measure of Angle A.
Recall that the sum of the interior angles of a triangle is 180°. This means:
[tex]AngleA+AngleB+AngleC=180\degree[/tex]Since we already know the measure of angles B and C, we can subtract it from the total measure which is 180° to solve for the measure of Angle A.
[tex]\begin{gathered} AngleA=180\degree-103\degree-28\degree \\ AngleA=49\degree \end{gathered}[/tex]Therefore, the measure of Angle A is 49°.
Next, let's solve for the measure of side c using Sine Law.
[tex]\frac{sinB}{b}=\frac{sinC}{c}[/tex]Let's plug into the equation above the value of B, C, and b.
[tex]\frac{sin103}{52}=\frac{sin28}{c}[/tex]Then, solve for the value of c.
Cross multiply.
[tex](c)(sin103)=(52)(sin28)[/tex]Divide both sides by sin 103.
[tex]\frac{(c)(sin103)}{(sin103)}=\frac{(52)(sin28)}{(sin103)}\Rightarrow c=\frac{52sin28}{sin103}[/tex]Evaluate the value of "c" using a calculator.
[tex]c=25.0546\approx25.1[/tex]The measure of side c is approximately 25.1 units.
Lastly, to solve for the measure of side "a", we will still use sine Law.
[tex]\frac{sinA}{a}=\frac{sinB}{b}[/tex]Let's plug into the equation above the value of A, B, and b.
[tex]\frac{sin49}{a}=\frac{sin103}{52}[/tex][tex]a=\frac{52sin49}{sin103}[/tex][tex]a=40.277\approx40.3[/tex]The measure of side a is approximately 40.3 units.