# How do you solve the right triangle ABC given B = 45 degrees, angle c = 95 degrees and side AB = 5 units?

May 17, 2015

Whoa! That's not a right triangle. But it can still be solved with the law of sines: $\sin \frac{A}{a} = \sin \frac{B}{b} = \sin \frac{C}{c}$ where A, B, and C are angles opposite sides a, b, and c, respectively.

Side AB is the only side given, so we'll work with that. This side actually represents side 'c' in the triangle, since it's opposite of angle C. It's also important that we also know angle A (40 degrees), because the angles in a triangle add up to 180 degrees.

$\sin \frac{{95}^{\circ}}{5} = \sin \frac{{45}^{\circ}}{A C} = \sin \frac{{40}^{\circ}}{B C}$

$0.19923893961 \approx \frac{\sqrt{2}}{2 A C} \approx \frac{0.64278760968}{B C}$

From here we split the equations.

$0.39847787922 \left(A C\right) \approx \sqrt{2}$

$A C = 3.54903907123$, or just 3.55 units.

$0.19923893961 \left(B C\right) \approx 0.64278760968$

$B C = 3.22621476975$, or just 3.23 units.