# How do you find the hypothesis of an obtuse and acute triangle given Base = 2 Vertical Side = 4, Angle = 75, which is opposite of the hypothesis, and Hypothesis = x?

Jul 13, 2018

$x = 2 \sqrt{5 - \sqrt{6} + \sqrt{2}}$

#### Explanation:

There are some things to clear up. The way you phrased your question, it was a bit confusing. First of all, a triangle cannot be obtuse and acute at the same time. In this particular case, the triangle is acute.

The side we wish to find the lenght; I do believe you meant hypotenuse, not hypothesis. However, only right triangles have hypotenuses, which are the sides opposite of the right angle. Also, it is somewhat incorrect to call the other sides "base" and "vertical side", due to this triangle not being right angled.

Anyway, here's our triangle:

Now, the Law of Cosines states that, if $a$ and $b$ are two sides of a triangle and the angle between them is $\theta$, then the third side, denoted $c$ here, is

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \theta$

$\therefore {x}^{2} = {2}^{2} + {4}^{2} - 2 \cdot 2 \cdot 4 \cos {75}^{\circ}$

We can calculate $\cos {75}^{\circ}$ using sum identities:

$\cos {75}^{\circ} = \cos \left({30}^{\circ} + {45}^{\circ}\right) = \cos {30}^{\circ} \cos {45}^{\circ} - \sin {30}^{\circ} \sin {45}^{\circ}$

$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$\therefore {x}^{2} = 4 + 16 - {\cancel{16}}^{4} \frac{\sqrt{6} - \sqrt{2}}{\cancel{4}}$

$\therefore {x}^{2} = 20 - 4 \sqrt{6} + 4 \sqrt{2}$

$\implies x = \sqrt{20 - 4 \sqrt{6} + 4 \sqrt{2}} = 2 \sqrt{5 - \sqrt{6} + \sqrt{2}}$