Q:

Find an angle in a triangle within a parallelogram

Someone sent me this exercise and I have been having trouble solving it. The exercise wants us to find theta, then tan(theta), and as you can see to the right, there are five alternatives and I tried everything I could think of but I don't get any of those five alternatives. I even tried using law of cosine and still got the same results, but I think I am doing something wrong. I inserted the angle theta that I got into tan but got none of the five alternatives. Any help appreciated!

A:

From a multiple-choice viewpoint, it's pretty clear that  $\theta>90°$ so $\tan \theta <0$.
For a specific calculation, the tangent addition formula is:
$$\tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$
So in this case we have (from your lower two triangle diagrams) that $\theta$ is composed of two angles that have tangent values of $\frac 52$ and $\frac 32$:
$$\tan \theta = \frac{\frac 52 +\frac 32}{1-\frac 52 \frac 32} = \frac{4}{\frac{4-15}{4}} = -\frac{16}{11}$$
Which is indeed none of the available choices. However there is something wrong with the rectangle in your first diagram - the top and bottom edges are different lengths - so that may be the issue.

