a = 15, b = 29 and c = 23
A= 31º B= 97º C= 52º
1) To solve a triangle is to find out its angles and measures. Since we have angle A= 31º, C =52º, and b= 29 we can solve this by using the Sum of Interior angles and the Law of sines:
[tex]\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}[/tex]2) The sum of the interior angles of a Triangle is 180º So, we can find angle B this way
∠A +∠B+∠C= 180º
31º +∠B +52º = 180º
∠B + 83º = 180º Subtract 83 from both sides
∠B =97º
[tex]\begin{gathered} \frac{29}{\sin(97)}=\frac{c}{\sin(52)} \\ \\ c=\frac{29\cdot\sin(52)}{\sin(97)}\Rightarrow c\approx23 \end{gathered}[/tex]2.2) Now we can find the leg "a "
[tex]\begin{gathered} \frac{a}{\sin (31)}=\frac{29}{\sin (97)} \\ a\cdot\sin (97)\text{ = 29 }\cdot\sin (31) \\ a=\frac{29\cdot\sin (31)}{\sin (97)}\Rightarrow a\approx15 \end{gathered}[/tex]3) Hence, the answer is
a = 15, b = 29 and c = 23
A= 31º B= 97º C= 52º