Trigonometry in Triangles Without Right-Angles Could you please help by showing me how I can find the unknown sides for the triangles below?



 A: Hint: since you do not have a right triangle, and do not know whether the second triangle is a right triangle, assume it isn't:
For the first, use the law of sines: the unknown angle is $180 - 60 - 50 = 70^\circ$
$$\frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C}\tag{Law of Sines}$$


For the second triangle: 
Use the law of cosines using the information you are given, to solve for the unknown, given the right choice of angles.
$$c^2 = a^2 + b^2 - 2ab\cos \gamma\tag{Law of Cosines}$$


Images from the linked wikipedia articles (respectively).
A: For the first one, use the fact that the sum of the interior angles of a triangle is $180^o$ to find the $3$rd angle and then use the sine law.
The third angle is $180-50-60=70$
The sine law states that ratio of the sines of two angles of a triangle is equal to the ratio of their opposite side lengths.
We can use it to find $b$: $\dfrac{b}{3}=\dfrac{\sin70}{\sin60}
\implies b\approx3.255$
For the second, use the cosine law using the formula provided by AmWhy. It's just the process of plugging in the values to yield the answer then.
