How do you solve the right triangle ABC given A= 20 degrees C= 104 degrees and c=9?

Jul 5, 2015

$\angle B = {180}^{o} - \left({20}^{o} + {102}^{o}\right) = {56}^{0}$
Using the Law of Sines
$\textcolor{w h i t e}{\text{XXXX}}$$a = 3.172416$
$\textcolor{w h i t e}{\text{XXXX}}$$b = 7.689757$

Explanation:

Ignoring the fact that the question declared that the triangle was a right triangle (obviously impossible with one angle = ${102}^{o}$):

The third angle can be calculated using the knowledge that the sum of the interior angles of any triangle is always ${180}^{o}$.

Using the Law of Sines in the form:
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{c}{\sin} \left(C\right) = \frac{a}{\sin} \left(A\right) = \frac{b}{\sin} \left(B\right)$
we can calculate (for example):
$\textcolor{w h i t e}{\text{XXXX}}$$a = \sin \left(A\right) \cdot \frac{c}{\sin} \left(C\right) = \sin \left({20}^{o}\right) \cdot \frac{9}{\sin} \left({102}^{o}\right)$
and through the magic of hand calculators or spreadsheet functions come up with the indicated answers.