So I understand that we can derive the degrees from a triangle with
$$TAN^{-1}\left(\frac{AB}{CB}\right)= A^o (degree)$$
But this only applies on a Right Triangle. In the next example I want to get value of $DF$ and all of the degrees in the triangle.
I can't seem to go any further, but I have done other Right Triangles loads of times.
This is what I have:
Split E so that we have 2 new smaller triangles, where EF and G (G is 90 degrees from E. Then DEG is on E:
$$140^0 - 90^0 = 50^0$$
And we know DE and EF. But from here it seems i'm lost.
For DF you can use cosine law: $$ \text{DF}²= \text{ED}² + \text{EF}² - 2\cdot\text{ED}\cdot\text{EF}\cdot\cos\text{DÊF} \\ \text{DF}²= 3² + 4² - 2\cdot3\cdot4\cdot\cos140^o $$
For the angles you can use sine law: $$ \frac{\text{DF}}{\sin140^o}=\frac{\text{ED}}{\sin\text{E}\hat{\text{F}}\text{D}}=\frac{\text{EF}}{\sin\text{E}\hat{\text{D}}\text{F}} $$