Hello, Bashyboy!

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

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

The ambiguity arises from using the inverse sine,

. . which always has two possible values: .

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.

We have: .

. .

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

But we already have two angles that total more than

Hence, the second (obtuse) triangle is impossible.

This is **not** the Ambiguous Case.

. . The only solution is *first* triangle.