01.24.2019: One Lesson of Math - The Sine Law, Part 2

Today's soundtrack is Triumph: Allied Forces, an album that I regret not giving its due up until recently. I got it with a few other CDs in one of those "mystery bag for $2" bins at Krazy Bob's Music Emporium back in the day, and because I'd bought it at a discounted rate, I unconsciously labeled it as a discount-level album. I've realized the error of my ways, though: it's a solid 80s rock album! It reminds me in a really good way of Journey's album "Eclipse," one of my favourite albums.

I learned how to use the sine law to determine the length of a side and to determine the measure of an angle in yesterday's lesson; today, I'm going to put those concepts together and learn how to use the sine law to solve a triangle.

Our first step is to determine how many possible triangles could be drawn. Based on this helpful explanation on socratic.com, I've put together a chart that shows the three possible outcomes of the six possible scenarios that we might run into when dealing with ambiguous SSA triangles. Note that h = b[sin A] (Citation).

Once we determine how many triangles there are, we must use the sine law to "write an equation that involves the two given lengths and the sines of their corresponding angles" (Pearson's Pre-Calculus 11, p. 475). Basically, we will be working with three fractions, of which after we will in what we know by using substitution, one will be entirely unknown, one will be half-known, and one will be entirely known. We need to start with the half-known and the entirely known, solve for the unknown angle in the half-known fraction, use the Triangle Sum Theorem to fill in the missing angle, and then solve for the missing side's length.

When we are writing out our sine law equations, we should put the unknown that we are solving for in the numerator position. We must also remember that when we are multiplying by an angle, 10(SIN5°) is not SIN50°, it is 10 SIN 5°.

If we know that the angle we are solving for is obtuse, after solving, we need to find its supplementary angle by subtracting our answer from 180°. The result will be our final answer.

Let's go through some examples...

If I was asked to find side b when I know that B = 30°, C = 80°, and c = 5cm, I would write out my equation like this: b / SIN 30° = 5 / SIN 80°. Then, using algebra to solve for b, I would multiply diagonally and divide vertically. So (SIN 30° ⋅ 5) / SIN 80°, which simplifies to 5 SIN 30° / SIN 80. 5 SIN 30° = 2.5; SIN 80° = 0.98. 2.5 / 0.98 = 2.54. This gives me my answer: b = 2.53cm.

In a problem where I need to find the measure of angle E when I know I know that side e = 8cm, d = 10cm, and D = 70°, I would start by writing out my equation: SIN E / 8 = SIN 70° / 10. I multiply diagonally and divide vertically: 8 (SIN 70°) / 10, which is 0.75. So we know that SIN E = 0.75; now, we need to find the measure of E. To do so, we calculate the inverse ratio to find the degrees: sin⁻¹ (0.75) = 48.59°. Let's say that we've been asked to find the degrees to the nearest tenth: our answer would be that E = 48.6°.

If I was asked to find angle J, which I know is obtuse, and I was told that j = 9cm, H = 50°, and h = 7cm, I would first create my equation: SIN J / 9 = SIN 50 / 7. Solving for SIN J, I would get the result 78.52°. But wait! 78.5° is not obtuse. So what gives? I need to find its supplementary. I subtract 78.5 from 180; the resulting answer, 101.5, is the solution. So J = 101.5°.

If given two sides and a non-enclosed angle, I can determine how many possible triangles could be constructed in the following way: I start by creating my equation in the same way as above based on the known information. For example, if I know that c = 10cm, a = 12 cm, and the angle of A = 20°, I will write SIN C / 10 = SIN 20° / 12. Using algebra, I can determine that C = 17°. Here's where the fun part comes in. Using the triangle sum theorem with my known angles (17° and 20°), I can say with certainty that the third angle is 143°. BUT - what if C is not actually acute? What if it's an obtuse angle? I'd better check to see whether that's a possibility. To find the supplementary of 17°, I subtract 17 from 180, which gives me 163. If I add that to our other known angle, 20°, I get 183°. Already I can see that this simply isn't possible! Every triangle has a total sum of 180°. So we can rule out the possibility of there being a second constructible triangle. If the supplementary of C plus the known angle had come to less than 180, there would be two possible triangles; if the angle of C was unsolvable (producing an error when trying to multiply the value by inverse sin), there would have been no possible triangles.