ambiguous case for law of sines

(Since there is no right angle, there is no hypotenuse )
 

A = 40°, B = 60°, c = 20

Let  a  be the length of the side opposite angle A.

Let  b  be the length of the side opposite angle B.

Let  C  be the measure of the angle opposite side c .

The sum of the angle measures in a triangle is  180°  , so...

A + B + C  =  180°

C  =  180° - A - B

C  =  180° - 40° - 60°

C  =  80°

Using the Law of Sines...

\(\begin{array}\ \frac{b}{\sin B}&=&\frac{c}{\sin C} \\ \frac{b}{\sin 60°}&=&\frac{20}{\sin 80°} \\ \quad b&=&\frac{20\sin60°}{\sin 80°} \\ \quad {\color{purple}b}&{\color{purple}\approx}&{\color{purple}17.6} \\~\\ \frac{a}{\sin A}&=&\frac{c}{\sin C} \\ \frac{a}{\sin 40°}&=&\frac{20}{\sin 80°} \\ \quad a&=&\frac{20\sin40°}{\sin 80°} \\ \quad {\color{purple}a}&{\color{purple}\approx}&{\color{purple}13.1 } \end{array}\)

(Also...this is not the ambiguous case...the ambiguous case is SSA)

Picture time! :D:D (I filled in B since \(180-(40+60) = 80\))

To solve for other sides, we can do the sine of the angles and add "x" whenever we need to solve for a variable.

\(sin(x)=\frac{opposite}{hypotenuse}\)

We have 20 units for c, so let's find a, since that's the opposite of A.

\(sin(A)=\frac{a}{c}\)

\(sin(40)=\frac{x}{20}\)

\(x = 20sin(40)\)

\(x = 14.90226320958​\)

\(x = 15\) \(units\)

Now that I solved for c, try and find b with the same method.

Alas as this is not a right triangle the basic trig ratios won't give you the correct answer. Here you need to employ the sin rule ie a/sinA = b/sinB = c/sinC