# 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!

• Why does side $AB$ have length $9$ while $DC$ only has length $8$? – WaveX Mar 24 '17 at 16:33
• Hint: $\tan \alpha_1$ and $\tan \beta_1$ can be easily calculated. – dxiv Mar 24 '17 at 16:34
• @WaveX Because the segment of length $3$ on $CD$ is longer than the segment of length $3$ on $AB$ ;-) – dxiv Mar 24 '17 at 16:36
• that is a typo, meant to write 2, it is 2 and not 3. Assume it is 3. Sorry, about that. – Dick Armstrong Mar 24 '17 at 16:42
• $\theta = 180^\circ - \alpha - \beta, \tan\theta = -\tan (\alpha + \beta) = -\frac {\tan \alpha +\tan\beta}{1-\tan\alpha\tan\beta}, \tan\alpha = \frac 25, \tan\beta = \frac 23$ – Doug M Mar 24 '17 at 16:43

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.

• I got the same answer with both cosine and sine laws, and the length issue in the diagram is just a mistake on my part, the 3 is supposed to be 2. Thank you for your answer! Now at least now I am not insane. – Dick Armstrong Mar 24 '17 at 17:02
• Now that I look more closely, I see that for all other measures it is a scale drawing, so I probably should have worked that out. – Joffan Mar 24 '17 at 17:04
• I have double checked my answers from all the methods so many times, I don't think I am wrong anymore, and I checked with you and others and every single person got the same results. I am certain the exercise is somehow messed up, at least their results don't match up with the actual exercise. I like your answer since it agrees with me and others I have spoken to and not to mention you have written it so well in my opinion, so I am accepting your solution as the answer. Thank you very much, Joffan! Much appreciated. – Dick Armstrong Mar 24 '17 at 19:13