# A triangle has sides A, B, and C. The angle between sides A and B is (3pi)/4. If side C has a length of 16  and the angle between sides B and C is pi/12, what are the lengths of sides A and B?

Feb 12, 2016

Side $b = 11.31$ and Side $a = 5.86$
:)

#### Explanation:

We can get the value of side A and B by using the "Law of Sines",

$\sin \frac{A}{a} = S \in \frac{B}{b} = S \in \frac{C}{c}$

First, we must convert the radian value to degree value.

To convert radian value to degree value,

Multiply it by $\frac{180}{\pi}$

since Angle $A = \frac{\pi}{12}$

$A = \frac{\pi}{12} \cdot \frac{180}{\pi}$ = $\frac{180 \pi}{12 \pi}$ = $\frac{180 \cancel{\pi}}{12 \cancel{\pi}}$

Angle $A = {15}^{o}$

and Angle $C = \frac{3 \pi}{4} \mathmr{and} {135}^{o}$

$C = \frac{3 \pi}{4} \cdot \frac{180}{\pi}$ = $\frac{540 \pi}{4 \pi}$ = $\frac{540 \cancel{\pi}}{4 \cancel{\pi}}$

Angle $C = {135}^{o}$

using the law of sines,

$\sin \frac{A}{a} = \sin \frac{C}{c}$

$\frac{\sin {15}^{o}}{a} = \frac{\sin {135}^{o}}{16}$

using algebraic technique we get,

$a = \frac{\left(\sin {15}^{o}\right) \left(16\right)}{\sin {135}^{o}}$

$a = 5.8564$

we use again the "Law of Sines", since "Pythagorean Theorem" doesn't work on Non-Right Triangles,

since angle $B$ is unknown, we can get its value by simply getting the difference of 180^o-(Angle$C +$Angle A), since "the sum of all interior angles of a triangle is always ${180}^{o}$"

Angle $B = {180}^{o} -$(135^o+15^o)

$= {180}^{o} - {150}^{o}$

Angle $B = {30}^{o}$

Applying again the Law of Sines to get the value of side of $b$, we get,

$\sin \frac{B}{b} = \sin \frac{C}{c}$

$\frac{\sin {30}^{o}}{b} = \frac{\sin {135}^{o}}{16}$

Applying algebraic technique, we get,

Side $b = \frac{\left(\sin {30}^{o}\right) \left(16\right)}{\sin {135}^{o}}$

Hence, we get:

Side $b = 11.3137$

Tip on Trigonometry:
Pythagorean Theorem is only reliable in solving right triangles, while Law of Sines and Cosines works in almost any triangles

:)