Ambiguous case using law of sines

Hello, here is the data I am given to work with:

Angle A=30 deg
Side a=3
Angle B=unknown
Side b=unknown
Angle C=unknown
Side c=4

So I know that I can take the 3/sin30 = 4/sinC
Rearranging that, I am left with SinC = Sin(30)(4)/3 which can further be simplified to 41.81 deg.

So does having two acute angles (30 deg. and 41.81 deg) account for the fact that it is ambiguous? Since we do not know the obtuse angle in the first place, which enables us to that think when we find out that angle C is acute, and we already have an acute angle, there is the possibility that it could actually be obtuse?

Also, I do not understand this part of the explanation in the book to this problem:

"Since angle C lies in a triangle a=30 deg. we must have that C greater than 0, but less than 150. There are two angles of C that fall in this range and have SinC = 2/3: C = (2/3) radians approx. 41.81 deg. and c = pi - arcsin (2/3) radians approx. 138.19 deg.

So I can comprehend the first sentence, but in the rest of the sentences I do not understand why they throw in pi and radians. I know pi is equal to 180 deg., but why did they throw in pi and radians so suddenly?

Absolutely first, you must determine if there is a triangle at all. Find the height of the triangle through B. h = 4*sin(30 deg) = 2. Since 3 > 2, we do have at least one triangle.

Second, how does side a compare to side c? Greater, and there is only one triangle, but two orientations. Less, and we have two.

Side 'a', then:

a < c*sin(A) there is no triangle. 'a' isn't long enough to read side 'b'.
a = c*sin(A) - One RIGHT triangle.
c < a < c*sin(A) there are two
a = c there is no triangle. 'a' overlaps 'c'.
a > c there is one triangle

If there are two, the construction of both with fixed A and c, creates an isosceles triangle with equal sides 'c'. If you construct this, you will see why the obtuse angle of triangle ABC is pi less acute angle C.

Hello, Bashyboy!

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

Quote:

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

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

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

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

No . . . all triangles have two acute angles.

The ambiguity arises from using the inverse sine,
. . which always has two possible values: . $\sin^{-1}\!\left(\tfrac{2}{3}\right) \:=\:\begin{Bmatrix}41.8^o \\ 138.2^o \end{Bmatrix}$
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.

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

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

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

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 $180^o.$

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

Hence, the second (obtuse) triangle is impossible.

This is not the Ambiguous Case.
. . The only solution is first triangle.