I attached a screen shot. Focus on the left triangle and the right triangle, not the whole thing. AD is in both triangles, and I made up a length of 10, AC looks a little bigger than AD, so I made up a length of 12. Since AC = x and AB = 2x, AB = 24. I used the law of cosines twice, once for AD and once for DC. Both of them are opposite 60 degree angles in triangles with a known side of 10.
Let AD be a, AC be b, and DC be c:
c^2 = a^2 + b^2  2ab cos (C)
cos (60) = 0.5, so 2 and cos (C) cancel out to make c^2 = a^2 + b^2  ab
c^2 = 10^2 + 12^2  10*12
c^2 = 100 + 144  120
c^2 = 124
I don't need to solve for c. For the other triangle where BD = 2(DC), c^2 = 4*124 = 496. That's not what I got below:
Let AD be a, AB be B, and BD be c:
c^2 = a^2 + b^2  2ab cos (C)
cos (60) = 0.5, so 2 and cos (C) cancel out to make c^2 = a^2 + b^2  ab
c^2 = 10^2 + 24^2  10*24
c^2 = 100 + 576  240
c^2 = 436
What did I do wrong?
Let AD be a, AC be b, and DC be c:
c^2 = a^2 + b^2  2ab cos (C)
cos (60) = 0.5, so 2 and cos (C) cancel out to make c^2 = a^2 + b^2  ab
c^2 = 10^2 + 12^2  10*12
c^2 = 100 + 144  120
c^2 = 124
I don't need to solve for c. For the other triangle where BD = 2(DC), c^2 = 4*124 = 496. That's not what I got below:
Let AD be a, AB be B, and BD be c:
c^2 = a^2 + b^2  2ab cos (C)
cos (60) = 0.5, so 2 and cos (C) cancel out to make c^2 = a^2 + b^2  ab
c^2 = 10^2 + 24^2  10*24
c^2 = 100 + 576  240
c^2 = 436
What did I do wrong?
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