Given:
In triangle [tex]ABC,m\angle A=138^\circ,c=19,a=29[/tex].
To find:
The [tex]m\angle C[/tex].
Solution:
According to the law of sines:
[tex]\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}[/tex]
Taking [tex]\dfrac{\sin A}{a}=\dfrac{\sin C}{c}[/tex], we get
[tex]\dfrac{\sin 138^\circ}{29}=\dfrac{\sin C}{19}[/tex]
[tex]\dfrac{0.66913}{29}\times 19=\sin C[/tex]
[tex]0.4383959=\sin C[/tex]
Taking sin inverse on both sides, we get
[tex]\sin{-1}0.4383959=C[/tex]
[tex]26.001577^\circ=C[/tex]
[tex]C\approx 26^\circ[/tex]
Therefore, the correct option is b.
