Hello, Bashyboy!

I have no idea why they threw in radians . . .

$\displaystyle \text{Given: }\;A=30^o,\;\;a=3,\;\; c = 4$

$\displaystyle \text{So I know that I can take: }\:\frac{3}{\sin30} \,=\, \frac{4}{\sin C}$

$\displaystyle \text{Rearranging, I am left with: }\:\sin C \:=\: \frac{4\sin30}{3} \:\approx\:41.81^o$

$\displaystyle \text{So does having two acute angles }(30^o,\:41.81^o)\text{ make it ambiguous?}$

No . . . *all* triangles have two acute angles.

The ambiguity arises from using the inverse sine,

. . which always has two possible values: .$\displaystyle \sin^{-1}\!\left(\tfrac{2}{3}\right) \:=\:\begin{Bmatrix}41.8^o \\ 138.2^o \end{Bmatrix}$

And sometimes the second angle is also a solution.

Sketch and label both triangles and see they are both possible.

Code:

B
*
* *
* 108.2 *
4 * *
* * 3
* *
* *
* 30 41.8 *
A * * * * * * * * * C

Code:

B
*
* *
* *
4 *11.8 *
* * 3
* *
* *
* 30 138.2 *
A * * * * * C

We see that both triangles are possible.

We have the Ambiguous Case.

How do we know if we do *not* have the Ambiguous Case?

There are various available tests,

. . but I have shown my students an elementary method.

$\displaystyle \text{Given: }\:A = 30^o,\;a = 6,\;c = 4$

We have: .$\displaystyle \frac{\sin C}{c} \:=\:\frac{\sin A}{a} \quad \Rightarrow\quad \sin C \:=\: \frac{c\sin A}{a} $

. . $\displaystyle \sin C \:=\:\frac{4\sin30^o}{6} \:=\:\frac{1}{3} \quad \Rightarrow\quad C \:=\:\sin^{-1}\!\left(\tfrac{1}{3}\right) \:\approx\:\begin{Bmatrix}19.5^o \\ 160.5^o\end{Bmatrix}$

Are both angles solutions?

Do we have the Ambiguous Case?

Sketch and label the triangles.

Code:

B
*
* *
* 130.5 *
4 * *
* * 3
* *
* *
* 30 19.5 *
A * * * * * * * * * C

This triangle is okay.

But this one is not . . .

Code:

B
*
* *
* ? *
4 * *
* * 3
* *
* *
* 30 160.5 *
A * * * * * C

The angles of a triangle must total $\displaystyle 180^o.$

But we already have two angles that total more than $\displaystyle 180^o.$

Hence, the second (obtuse) triangle is impossible.

This is **not** the Ambiguous Case.

. . The only solution is *first* triangle.