Michael B. | I can provide your 'A-HA' momentI can provide your 'A-HA' moment
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Maria,
as you say, there is not one unique solution for this triangle. Using Law of Sines, we get:
sin C = 18 * sin(50) / 14 = 0.98491
There are two angles which satisfy this constraint: C = 80.035 and C = 180 - 80.035 = 99.965
From these two answers we can derive the rest of the triangle by summing the angles to 180 degrees and then using Law of Sines to solve for the remaining side:
A = 49.965
B = 50
C = 80.035
a = 13.993
b = 14
c = 18
or
A = 30.035
B = 50
C = 99.965
a = 9.148
b = 14
c = 18
The reason there are actually two different possible answers is that you are given 2 sides of the triangle, but the angle you're given isn't between those sides. This is exactly why you can't prove triangle congruence by using Angle, Side, Side (also, the acronym is naughty, haha).
To get both possible solutions, use the Law of Cosines (c2 = a2 + b2 - 2ab*cos(C)). Since you know 2 sides and an angle, set it up so that the only variable is the missing side; to make the letters match, use the version of the Law of Cosines that looks like "b2 = a2 + c2 - 2ac*cos(B)". This will give you a quadratic: 196 = 324 + x2 -2*18*x*cos(50). Make one side equal zero by subtracting the 196 over to the other side and use the quadratic function (x = (-b ± √(b^2 - 4ac))/(2a)) to get both answers. I'll leave that part to you.
To be clear, it isn't always the case that SSA produces two distinct solutions - there may be two, one, or zero. The case of one distinct solution necessarily turns out to be a right triangle.
True. I just wanted to relate what was (hopefully) existing knowledge about proving triangle congruence to this problem (forming links to previously understood topics helps retention). Also, in my experience, it's easier for students to see a quadratic and know there are 0, 1, or 2 solutions (as you mentioned) than for them to see sin T = x and remember there might be more than one solution for T.
Yeah it's definitely a good way to prove to yourself that there should be two answers (in this case). Your experience is different from mine, though - I've never had one student get a visceral understanding using your method - perhaps this is my own failing. That's why I wish so much that Wyzant would give us some way to post images in our responses. That's been the best method for my students with this type of problem - showing how that "third side" can swing into two different positions for the given constraints if it's long enough, 1 position if it's just the right length, and none if it is too short. That visualization is something that usually sticks with them very well. Maybe I'll turn it into a blog post on my non-Wyzant blog.
One way to solve this problem is to use the Law of Sines: sin(A)/a = sin(B)/b = sin(C)/c. Here a, b, c are the lengths of the sides, and A, B, C are the angles across from (not touching) the sides a, b, c respectively.
Since you are given b, c, and B, you could plug them into sin(B)/b = sin(C)/c to find C. Having found C, use the fact that the angles of a triangle sum to 180 degrees to find A. Then plug A, B, and b into sin(A)/a = sin(B)/b to find a.
Comments
To be clear, it isn't always the case that SSA produces two distinct solutions - there may be two, one, or zero. The case of one distinct solution necessarily turns out to be a right triangle.
True. I just wanted to relate what was (hopefully) existing knowledge about proving triangle congruence to this problem (forming links to previously understood topics helps retention). Also, in my experience, it's easier for students to see a quadratic and know there are 0, 1, or 2 solutions (as you mentioned) than for them to see sin T = x and remember there might be more than one solution for T.
Yeah it's definitely a good way to prove to yourself that there should be two answers (in this case). Your experience is different from mine, though - I've never had one student get a visceral understanding using your method - perhaps this is my own failing. That's why I wish so much that Wyzant would give us some way to post images in our responses. That's been the best method for my students with this type of problem - showing how that "third side" can swing into two different positions for the given constraints if it's long enough, 1 position if it's just the right length, and none if it is too short. That visualization is something that usually sticks with them very well. Maybe I'll turn it into a blog post on my non-Wyzant blog.