Determine whether the given information results in one triangle, two triangles, or no triangle at all. Solve any resulting triangle(s).

a = 3

b = 2

A= 50^{o}

AdamTaurus May 16, 2019

#1**+1 **

Plane geometry uses a postulate called "side, side, angle" to prove the congruency of two triangles. What that means is that if you have three parameters of a triangle, as long as at least one of them is a side, it can be only one triangle. The information you have there can form only a single triangle.

ssa

Guest May 16, 2019

#2**+3 **

Guest, it is not true that we can always form a triangle given 2 sides and a non-included angle.

For example, let a = 1, b = 2, A = 45°

The length of side c is not fixed and the measure of angle C is not fixed.

There is still no way to exend/shorten c or rotate a so that a touches c .

Side a is just too short to ever touch side c .

It is also possible that two different triangles can be formed from the same given side-side-angle.

For an example of that, see: http://www.softschools.com/math/calculus/the_ambiguous_case_of_the_law_of_sines/

hectictar May 16, 2019

#4**+1 **

Hectictar, I see what you mean. For this specific problem, I was visualizing a triangle similar to the obtuse triangle in the example you linked to. Such a triangle is possible to form using the measurements provided. I neglected to consider whether the same triangle located in a different quadrant could be considered a separate triangle. Thanks for the clarification.

.

Guest May 16, 2019

#3**+3 **

We can use the law of sines:

\(\begin{array}{rcl} \frac{\sin A}{a}&\,=\,&\frac{\sin B}{b}\\~\\ \frac{\sin(50°)}{3}&\,=\,&\frac{\sin B}{2}\\~\\ 2\cdot\frac{\sin(50°)}{3}&\,=\,&\sin B\\~\\ \frac{2}{3}\sin(50°)&=\,&\sin B \end{array} \\ \ \\~\\ B\approx30.71°\quad\text{or}\quad B\approx149.29° \)

There are infinitely many values of B that produce the same sin value, but we only want to consider those

that are in the interval (0, 180°) because no angle in a triangle can have a measure outside that interval.

This is how we find those two solutions:

\(B\,=\,\arcsin(\frac{2}{3}\sin(50°))\qquad\text{or}\qquad B\,=\,180°-\arcsin(\frac{2}{3}\sin(50°))\)

But in this case, B ≈ 149.29° can't be a solution because we already have an angle that is 50°, and

149.29° + 50° > 180°

So we know B ≈ 30.71° is the only solution for this problem. So there is only one possible triangle.

Now we can find the measure of angle C .

C = 180° - A - B ≈ 180° - 50° - 30.71° ≈ 99.29°

All we are missing is the length of side c. We can find it using the law of cosines.

c^{2} = a^{2} + b^{2} - 2ab cos C

c^{2} ≈ 3^{2} + 2^{2} - 2(3)(2) cos( 99.29° )

c^{2} ≈ 13 - 12 cos( 99.29° ) c is a length, so take the positive sqrt of both sides

c ≈ √[ 13 - 12 cos( 99.29° ) ] Plug this into a calculator

c ≈ 3.865

Now we have found:

B ≈ 30.71°

C ≈ 99.29°

c ≈ 3.865

hectictar May 16, 2019