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# Determine

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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= 50o

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

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/

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
#5
+3 Notice that in both of these triangles,

AC  =  10

AB'  =  AB  =  6

m∠ACB  =  33°

However, these two are not the same triangle because

CB'  ≠  CB

m∠AB'C  ≠  m∠ABC

m∠CAB'  ≠  m∠CAB

hectictar  May 16, 2019
edited by hectictar  May 16, 2019
edited by hectictar  Jun 7, 2019
#6
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Yes, I get it now.

Thanks again.

Guest May 17, 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.

c2  =  a2 + b2 - 2ab cos C

c2  ≈  32 + 22 - 2(3)(2) cos( 99.29° )

c2  ≈  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

May 16, 2019