# Finding Angles from Side Length in Pre-Calculus

I had this problem on a recently college exam. I had no idea how to do it, and lost all my points for it. I assumed it would have something to do with finding the correct values for the sides and angles of a triangle given the sine and cosine rules, but the problem didn't seem to give enough space to solve them with that.

On my exam my professor wrote the tangent addition identity:

$$\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A) \cdot \tan(B)}$$

But I'm not sure what relevance this has.

If C = A + B, find tan(C) • Are the two angles on the top triangles right angles? – KM101 Nov 21 '18 at 20:45
• @KM101 Yes, that's correct. – LuminousNutria Nov 21 '18 at 20:46
• There is no need for anything really. You’re given the opposite and adjacent sides to $\angle A$ and $\angle B$, so you can easily find $\tan A$ and $\tan B$. From there, you use the identity. – KM101 Nov 21 '18 at 20:53

Hint: $$\tan A = \frac{1}{2}$$

$$\tan B = \frac{3}{4}$$

Now, apply $$\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}$$

• How do you know the value of $\tan A$ and $\tan B$ here? How do you know that the botton side of those triangles is the hypotenuse? – LuminousNutria Nov 21 '18 at 20:54
• The bottom side IS the hypotenuse, so the sides given by the question are the opposite and adjacent sides. – KM101 Nov 21 '18 at 20:55
• Sorry, I just edited my post. How do you know which side is the hypotenuse? – LuminousNutria Nov 21 '18 at 20:56
• It seems to be implied by the “squar-ish” angle marks. – KM101 Nov 21 '18 at 20:58
• Oh, right. Since the sum of the angles of a triangle is 180 degrees, and 90 degrees is half of that, the other two angles must be smaller. Since the other angles are smaller, the sides opposite them must also be smaller. Therefore the side opposite a 90 degree angle in a triangle is always the hypotenuse. Thanks! – LuminousNutria Nov 21 '18 at 21:00

$$\tan(C) =\tan(A+B) = \frac{1/2 + 3/4}{1 - 3/8} = \frac{5/8}{5/8} = 1$$