Question

Solve the triangle: a = 12,c = 2-2, B = 33". If it is not possible, say so.A= 25.1",b = 1.8, C = 121.9"This triangle is not solvable.A = 45*,b= V2.C = 102VEA= 30', b = -, C = 117"

Answer 1

ANSWER:

A=25.1 degrees

b = 1.8

C = 121.9 degrees

SOLUTION:

We can solve this problem using the cosine law, since we are given the length of 2 sides of triangle and the angle they formed.

$b\text{ =}\sqrt[]{c^2+a^2-2ac\cos B}$

We substitute the given

$\begin{gathered} b\text{ =}\sqrt[]{(2\sqrt[]{2})^2+(\sqrt[]{2})^2-2(\sqrt[]{2})(2\sqrt[]{2})\cos 33} \\ b\text{ = 1.8} \end{gathered}$

Using Sine Law, we can get the angles

$\begin{gathered} \frac{1.8}{\sin 33}=\frac{\sqrt[]{2}}{\sin A} \\ A=25.1 \end{gathered}$

Since the total angle inside a triangle is 180, the angle at C is

$C-33-25.1=121.9$