Given the triangle ABC and knowing that:
[tex]\begin{gathered} a=BC=15 \\ c=AB=17 \\ m\angle B=137º \end{gathered}[/tex]You need to apply:
• The Law of Sines in order to solve the exercise:
[tex]\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}[/tex]That can also be written as:
[tex]\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}[/tex]Where A, B, and C are angles and "a", "b" and "c" are sides of the triangle.
• The Law of Cosines:
[tex]b=\sqrt[]{a^2+c^2-2ac\cdot cos(B)}[/tex]Where "a", "b", and "c" are the sides of the triangle and "B" is the angle opposite side B.
Therefore, to find the length "b" you only need to substitute values into the formula of Law of Cosines and evaluate:
[tex]b=\sqrt[]{(15)^2+(17)^2-2(15)(17)\cdot cos(137\degree)}[/tex][tex]b\approx29.8[/tex]• To find the measure of angle A, you need to set up the following equation:
[tex]\begin{gathered} \frac{a}{sinA}=\frac{b}{sinB} \\ \\ \frac{15}{sinA}=\frac{29.8}{sin(137\degree)} \end{gathered}[/tex]Now you can solve for angle A. Remember to use the Inverse Trigonometric Function "Arcsine". Then:
[tex]\frac{15}{sinA}\cdot sin(137\degree)=29.8[/tex][tex]\begin{gathered} 15\cdot sin(137\degree)=29.8\cdot sinA \\ \\ \frac{15\cdot sin(137\degree)}{29.8}=sinA \end{gathered}[/tex][tex]A=\sin ^{-1}(\frac{15\cdot sin(137\degree)}{29.8})[/tex][tex]m\angle A=20.1\degree[/tex]• In order to find the measure of Angle C, you need to remember that the sum of the interior angles of a triangle is 180 degrees. Therefore:
[tex]m\angle C=180º-137º-20.1\degree[/tex]Solving the Addition, you get:
[tex]\begin{gathered} m\angle C=180º-137º-20.1\degree \\ m\angle C=22.9\degree \end{gathered}[/tex]Therefore, the answer is:
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