To solve this question we will use the sine Law:
[tex]\frac{c}{\sin C}=\frac{b}{\sin B}=\frac{a}{\sin A}\text{.}[/tex]To find the measure of angle A, recall that the interior angles of a triangle add up to 180 degrees, therefore:
[tex]\measuredangle A+\measuredangle B+\measuredangle C=180^{\circ}.[/tex]From the given data we know that:
[tex]\begin{gathered} \measuredangle B=19^{\circ}, \\ \measuredangle C=66^{\circ}. \end{gathered}[/tex]Therefore:
[tex]\measuredangle A+19^{\circ}+66^{\circ}=180^{\circ}.[/tex]Solving the above equation for angle A we get:
[tex]\begin{gathered} \measuredangle A=180^{\circ}-19^{\circ}-66^{\circ}, \\ \measuredangle A=96^{\circ}. \end{gathered}[/tex]Now, from the given data we get that b=5, therefore:
[tex]\begin{gathered} \frac{a}{\sin96^{\circ}}=\frac{5}{\sin 19^{\circ}}, \\ \frac{c}{\sin 66^{\circ}}=\frac{5}{\sin19^{\circ}}\text{.} \end{gathered}[/tex]Solving the above equations for a and c respectively we get:
[tex]\begin{gathered} \frac{a}{\sin96^{\circ}}\times\sin 96^{\circ}=\frac{5}{\sin19^{\circ}}\times\sin 96^{\circ}, \\ a=\frac{5\sin96^{\circ}}{\sin19^{\circ}}, \\ \frac{c}{\sin66^{\circ}}\times\sin 66^{\circ}=\frac{5}{\sin19^{\circ}}\times\sin 66^{\circ} \\ c=\frac{5\sin66^{\circ}}{\sin19^{\circ}}, \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} a\approx15.3, \\ c\approx14.0. \end{gathered}[/tex]Answer:
[tex]\begin{gathered} a\approx15.3, \\ c\approx14.0. \end{gathered}[/tex]
