Scribe Post review 5.1-5.3 Nigel.

So in class tuesday we looked over some problems that could possible be on the test for a while then we had time to work on our own once we asked all of our questions.

The ambiguous case
The so-called ambiguous case arises from the fact that an acute angle and an obtuse angle have the same sine. If we had to solve
sin x = ½,
for example, we would have
x = 45° or x = 135°.
(Topic 6, Example 1.)
In the following example, we will see how this ambiguity could arise.

In triangle ABC, angle A = 30°, side a = 1.5 cm, and side b = 2 cm. Let us use the law of sines to find angle B.
sin B
sin 30° = 2
1.5

Since sin 30° = ½ (Topic 7, Example 2),
sin B = ½· 20
15
= 10
15
= 2
3
.666
On inspecting the Table for the angle whose sine is closest to .666, we find
B42°.
But the sine of an angle is equal to the sine of its supplement. That is, .666 is also the sine of 180° − 42° = 138°.
This problem has two solutions. Not only is angle CBA a solution,

but so is angle CB'A, which is the supplement of angle CBA. (We can see that it is the supplement by looking at the isosceles triangle CB'B; angle CB'A is the supplement of angle CB'B, which is equal to angle CBA.)
Given two sides of a triangle a, b, then, and the acute angle opposite one of them, say angle A, under what conditions will the triangle have two solutions, only one solution, or no solution?

Let us first consider the case a < b. Upon applying the law of sines, we arrive at this equation:
1) sin B = sin A· b
a .
Now, since h
b = sin A , where h is the height of the triangle (Fig. 1),
then
b sin A = h.
On replacing this in the right-hand side of equation 1), it becomes
sin B = h
a .
There are now three possibilities:

h
a < 1, which implies h < a (Fig. 1),
h
a = 1, which implies h = a (Fig. 2),
h
a > 1, which implies h > a (Fig. 3).
In the first of these -- h or b sin A < a -- there will be two triangles.
In the second -- h or b sin A = a -- there will be one right-angled triangle.
And in the third -- h or b sin A > a -- there will be no solution.