I’m studying for the SAT Math II test with Barron’s and their explanation for the ambiguous case in trigonometry is very bare. Here’s the question I’m stuck on (page 79, question 13):
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’ve drawn a dozen triangles and still can’t understand Barron’s explanation, which goes: In I, the altitude = 12 * 0.5 = 6,
6 < c < 12, and so 2 triangles. In II, b > 12sqrt(2), so only 1 triangle. In III, the altitude = 12 * (sqrt(3)/2) > 5sqrt(3), so no triangles.
I just can’t seem to wrap my head around the visualization of where to place the points and angles, and every time I draw the problem it seems to be different. Please help!
I can confirm the answer is correct, although yes, the explanation can be a little tricky without good drawing skills or a diagram.
By convention, side c is opposite angle C, and so on. For I., I would draw a 30 degree angle (label it C), then a segment of length 12 (label it “b”) with one endpoint at C. (orientation doesn’t matter since the worst you get is a reflection). Then the segment’s other endpoint is A.
The altitude from A to BC is 6, which is what the explanation says (although the explanation is poorly worded). Because c = 8 is greater than the altitude and less than the length of b, two non-congruent triangles can be formed using the givens.
In general, I refer my kid to what she has learned…In this case, I think the ambiguous case in law of sines applies.
https://en.wikipedia.org/wiki/Law_of_sines
something she learned in pre-cal