I’ve been going through my Barron’s Math 2 book and holy crap, there’s a ton of bad typos in this thing, but I’m still trying to make the most of it. This is the 11th edition on page 75 for anyone that has it. But what it asks is:
Given the following data, which can form 2 triangles?:
I. Angle C = 30°, c=8, b=12
II. Angle B = 45°, a =12sqrt(2), b = 15sqrt(2)
III. Angle C = 60°, b=12, c=5*sqrt(3)
I really can’t quite grasp this question and the explanation Barron’s provides isn’t quite clear to me:
For example Barron’s says for I:
In I the altitude = 12 * 1/2 = 6 and 6 < c < 12, and so two triangles
However I don’t know why you’d do 12 * 1/2 (which is 12sin30) because wouldn’t that just give you side c again?
Their explanation seems poor, but I think what they’re trying to get at is that the altitude from A to BC (opp. the 30 deg angle) is 6, and because c (side AB) is strictly between 6 and 12, you can draw two noncongruent triangles. Try constructing it on paper.
In general, knowing two sides of a triangle with the angle in between is enough to uniquely determine the triangle, but two sides with either of the other angles usually isn’t enough (hence no SSA congruence).
@MITer94 Oh, I’ve drawn the triangle probably like 6 times trying to understand it. The only way that seems to make sensr to me is going through the full law of sines process by first finding angle B and then doing 180° - (angle b + angle c) to find angle A and then just realizing that 180 - 101.4° (the resulting angle) can also be added to 30° without going over 180°.
I tried to figure out how to find the altitude from angle A to BC but I don’t know how without finding angle A first and finding the length of side a (BC).
@bringit1 here is a sketch of what I looks like (excuse my bad MS paint skills 
)
http://imgur.com/PJfOjLU (imgur)
Basically, you are given the 30° angle and b = 12, so I constructed that first. Point B satisfies AB (or side c) = 8, so B can be at either B1 or B2 (as shown). The altitude from A to BC must be 6, because 12 sin 30° = 6. Because 6 < c < 12, that’s why you get two possible points.
Also note that I didn’t need to find the length of BC or the angle measure of A because that’s irrelevant.
@MITer94 See, I feel like an idiot because what I’m not understanding is how you get that altitude of 6. Shouldn’t that altitude go from B1 or B2 straight down to side B? Also, I thought one triangle would be obtuse (101.4°) and the other acute with an angle of 78.6° instead.
@bringit1 no, the altitude I drew goes from A to the midpoint of B1 and B2 (so that a right angle is formed). Consider only the right triangle formed with the right angle and hypotenuse b = 12.
The obtuse triangle has angle measure 180° - sin^(-1) (6/8) ≈ 131.4° but that fact isn’t needed.
Also my drawing isn’t to scale; triangle AB_2 C is actually obtuse…
Ah, I see. I didn’t know you use the altitude from the triangle formed by the “difference” between the two triangles. Barron’s really doesn’t seem to clarify such concepts but that’s fine I guess. Just using the Law of Sines seems much more simple and easy to remember for me though. I don’t think there’d be a case where just calculating the angle wouldn’t work when this method would, is there?
Doubt it, but in this problem you don’t need to calculate any angles.