Question

Help please ! thank u

Answer 1

Given:

In triangle $ABC,m\angle A=138^\circ,c=19,a=29$.

To find:

The $m\angle C$.

Solution:

According to the law of sines:

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

Taking $\dfrac{\sin A}{a}=\dfrac{\sin C}{c}$, we get

$\dfrac{\sin 138^\circ}{29}=\dfrac{\sin C}{19}$

$\dfrac{0.66913}{29}\times 19=\sin C$

$0.4383959=\sin C$

Taking sin inverse on both sides, we get

$\sin{-1}0.4383959=C$

$26.001577^\circ=C$

$C\approx 26^\circ$

Therefore, the correct option is b.