Given a triangle with a =19, A = 43°, and B =26°, what is the length of c? Round to the nearest tenth

a = 19

A = 43°

B = 26°

 

 

I know the problem says to find the length of c; however, angle C and length b can also be found using the Law of Sines and that a traingle is equal to 180°.

 

\(\frac{sin(A)}{a}=\frac{sin(B)}{b}=\frac{sin(C)}{c}\)

 

To find angle C, subtract angles A and B from 180°.

 

\(C=180°-43°-26°\)

 

\(C=137°-26°\)

 

\(C=111°\)

 

Now I will find length b.

 

\(\frac{sin(43°)}{19}=\frac{sin(26°)}{b}\)

 

\(19b\times\frac{sin(43°)}{19}=19b\times\frac{sin(26°)}{b}\)

 

\(sin(43°)b=19sin(26°)\)

 

\(\frac{sin(43°)b}{sin(43°)}=\frac{19sin(26°)}{sin(43°)}\)

 

\(b=\frac{19sin(26°)}{sin(43°)}\)

 

\(b≈\frac{19\times0.438371146789}{sin(43°)}\)

 

\(b≈\frac{8.329051788991}{sin(43°)}\)

 

\(b≈\frac{8.329051788991}{0.681998360062}\)

 

\(b≈12.2127152743209113\)

 

\(b≈12.2\)

 

Now I will length c.

 

\(\frac{sin(43°)}{19}=\frac{sin(111°)}{c}\)

 

\(19c\times\frac{sin(43°)}{19}=19c\times\frac{sin(111°)}{c}\)

 

\(sin(43°)c=19sin(111°)\)

 

\(\frac{sin(43°)c}{sin(43°)}=\frac{19sin(111°)}{sin(43°)}\)

 

\(c=\frac{19sin(111°)}{sin(43°)}\)

 

\(c≈\frac{19\times0.933580426497}{sin(43°)}\)

 

\(c≈\frac{17.738028103443}{sin(43°)}\)

 

\(c≈\frac{17.738028103443}{0.681998360062}\)

 

\(c≈ 26.0089014023881936\)

 

\(c≈26.0\)

 

A = 43°

B = 26°

C = 111°

a = 19

b ≈ 12.2

c ≈ 26.0