 In this video, we provide the solution to question number 20 from the practice final exam for math 1060, in which case we're given a triangle for which one of the interior angles is 45 degrees. Its opposite side is the square root of two centimeters. And then one of the sides next to the angle is one centimeter, and we're supposed to solve for this triangle if possible. Well, since we have angle A and side length A, we have this angle opposite side pair. I'm gonna try to use the law of signs to solve for this triangle right here. So my target is gonna go after angle B right there. So by the law of signs, we get that sine of B over one is equal to sine of 45 degrees over the square root of two. For which simplifying that, of course, we're just gonna end up with sine of B, which is equal to, well, sine of 45 degrees is root two over two. You divide that by the square root of two, which to help you out here, you can think of this as just multiplying by the reciprocal, right? So the square roots of two cancel out and we end up with sine of B is equal to one half. And so taking arc sine of that, if you just use your calculator, your calculator will report back 30 degrees right here. But one has to be careful, right? Because when you take sine inverse of one half, 30 degrees of course is correct, but we have to also remember that sine can't tell the difference between an acute angle and obtuse angle. That is to say, since sine is positive in the first quadrant and in the second quadrant, we have to consider both possibilities. There's the 30 degrees, but there's also its supplement 150 degrees. Cause notice right now we're in the ambiguous case, this side-side angle situation. So we have to consider the two possibilities. So what's the first possibility, right? The first triangle we're taking as our situation 30 degrees for B, then what would C be? Angle C would be 180 degrees, takeaway B, which is 30 degrees, takeaway A, which was 45 degrees. So we have 180 takeaways, 75 degrees. We end up with C, then being 105 degrees, like so. And so then we have to find the third side of the triangle. We could do that using the law of sines or the law of cosines, whichever you prefer. I'm just gonna use the law of sines in this situation. So we get that little C over sine of 105 degrees. This is the same thing as, I'm just gonna use the original side. We have square root of two over sine of 45 degrees, like so. For which that tells us that C, excuse me, C is equal to the square root of two times sine of 105 degrees over root two over two. Which again, as we're dividing by a fraction, this is the same thing as having, think of it this way, we're gonna take two over the square root of two, like so, so those cancel out. And so we see, look at C here, it's two times sine of 105 degrees. 105 degrees of course is an obtuse angle. Sine doesn't tell the difference between obtuse angle and acute angle. So this actually would be the same thing as its reference angle, two times sine of 75 degrees. For which you might remember sine of 75 degrees is equal to the square root of six plus the square root of two over four. And so simplifying, you end up with the square root of six plus the square root of two over two. Now, if you instead get an approximate answer here because you don't remember the exact value of 105, sine of 105 or 75, that's perfectly acceptable, no big deal. And so this gives us one triangle. So one triangle is when B equals 30 degrees, C equals 105 degrees, and then the side length C is the square root of six plus the square root of two over two or approximately, again, if you did an approximate answer, you get something like 1.932, like so. What about the second triangle though? There is a second possibility. What happens if B equals 150 degrees? Well, that would then suggest that C equals 180 degrees minus B, which is 150, minus A, which is 45, that actually ends up with negative 15 degrees for C, which that's not a possibility. And so we actually see that the second triangle is not possible. There's only gonna be the one triangle, the one we found earlier with 30 degrees, 105 degrees and 45 degrees. Now in the ambiguous case, you do have to check this for the second triangle. Because sometimes it works, sometimes it doesn't. And so for full credit on this one, if you don't check for the second triangle, you wouldn't get full credit. So you have to check even if it turns out the second triangle doesn't exist.