The side lengths of a cyclic quadrilateral ABCD are provided in the diagram. Find BD^2.
The picture is here as I can't upload it for some reason: https://cutt.ly/auSLG9T
I think there might be some "formula" for solving this in an easier fashion, but, I believe the Law of Cosines might also work
Note that angles DCB and DAB are supplemental so cos ( DCB) = -cos (DAB)
Using the Law of Cosines we have this system (and subbing -cos (DAB) for cos (DCB) )
DB^2 = 3^2 + 2^2 - 2 ( 3 * 2) (-cos DAB)
DB^2 = 1^2 + 4^2 - 2 ( 4 * 1) (cos DAB)
DB^2 = 13 + 12 cos DAB
DB^2 = 17 - 8 cos DAB
Multiply the first equation by 8 and the second by 12
8 DB^2 = 104 + 96 cos DAB
12 DB^2 = 204 - 96 cos DAB add these
20 DB^2 = 308
DB^2 = 308 / 20 = 15.4
The side lengths of a cyclic quadrilateral ABCD are provided in the diagram. Find BD^2.
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a = 1 b = 2 c = 3 d = 4 s = 5
Using the above formula, we can calculate the circumradius R.
R = 2.002602473
Now, we can calculate the measure of angle C.
Angle C = 101.53695902º
Finally, using the law of cosines, we can find the length of BD.
BD ≈ 3.924 BD2 = 15.4