# How do you solve a triangle given <A=84.2°, <B=20.7°, B=17.2?

Apr 24, 2016

The following diagram represents your problem

#### Explanation:

Step 1:

Since we know an angle opposite a known side, we can use the Sine Rule to solve for side a.

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

$\sin \frac{84.2}{a} = \sin \frac{20.7}{17.2} = \sin \frac{C}{c}$

$a = \frac{17.2 \times \sin 84.2}{\sin} 20.7$

$a = 48.4$

Step 2:

We could have done this at first, but let's find angle C. We know that the sum of the three angles in a triangle always equals 180. Therefore, we can set up the following equation:

$C + 20.7 + 84.2 = 180$

$C = 180 - 20.7 - 84.2$

C = 75.1˚

Step 3:

We must now substitute the value of angle C to find side C.

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

$\sin \frac{20.7}{17.2} = \sin \frac{75.1}{c}$

$\frac{17.2 \times \sin 75.1}{\sin} 20.7 = c$

$47.0 = c$

Summary:

Your triangle has the following dimensions:

/_A = 84.2˚
/_B = 20.7˚
/_C = 75.1˚

$a = 48.4$ units
$b = 17.2$ units
$c = 47.0$ units

Hopefully this helps!