This might be fairly easy to some.

Test Preparation
1

<li>How many distinct triangles can be formed if the measure of angle A =30, side b = 12, and side a = 8?</li>
</ol>

<p>(1) 1
(2) 2
(3) 3
(4) 0</p>

<p>Explain please.</p>

2

<p>You have to use the law of sines.</p>

<p>sin A / a = sin B / b</p>

<p>sin 30 / 8 = sin B / 12</p>

<p>.5 / 8 = sin B / 12</p>

<p>6 = 8 sin B
6/8 = sin B
3/4 = sin B</p>

<p>sin^-1(.75) = ~48.59 deg</p>

<p>sin is positive in the II quadrant...so
180 - 48.59 = 131.41</p>

<p>so B can be either 48.59 or 131.41 deg...I think you have to verify it by making sure that everything else works...I forgot all this stuff.</p>

<p>triangle 1:
A = 30
B = 48.59
C = 101.41
a = 8
b = 12
c =</p>

<p>sin A /a = sin C / c
sin 30 / 8 = sin 101.41 / c</p>

<p>.0625 = .9802 / c
.0625c = .9802
c = 15.68</p>

<p>triangle 2
A = 30
B = 131.41
C = 18.6
a = 8
b = 12
c =</p>

<p>sin A /a = sin C / c
sin 30 / 8 = sin 18.6 / c
.0625 = .3190 /c
.0625c = .3190
c = .1959</p>

<p>draw the triangles...does it make sense? yeah it does (use the triangle inequality thm...or w/e its called)</p>

<p>so yeah...i would pick 2 different triangles</p>

3

<p>Oh. Thanks. I forgot that there were two values for sine.</p>

4

<p>No problemo.</p>

5

<p>If you just know that the information given requires the Law of Sines to solve, and you know that the Ambiguous Case can give you two different triangles, do you really need to solve the problem?</p>

6

<p>Umm....I want to say you don't have to solve it...but my gut says you do...b/c I think there are some cases where the second triangle doesn't work b/c of the triangle inequality thm (or w/e its called).</p>

7

<p>Gut got you good. :D</p>

<p>Let's look at the right triangle ABC: A = 30, B = 90, b = 12.
Since CB = 6, on line AB (extended past point B) there are two points, B1 and B2, symmetrical about point C, such that CB1 = CB2 = 8.
Both triangles ACB1 and ACB2 have A = 30, b = 12, and a = 8.</p>

<p>The answer: two distinct triangles.</p>

8

<p>Let's look at the right triangle ABC: A = 30, B = 90, b = 12.
Since CB = 6, on line AB (extended past point B) there are two points, B1 and B2, symmetrical about point B, such that CB1 = CB2 = 8.
Both triangles ACB1 and ACB2 have A = 30, b = 12, and a = 8.</p>

<p>The answer: two distinct triangles.</p>