In the diagram, \(AB = 13\text{ cm},\) \( DC = 20\text{ cm},\) and \( AD = 5\text{ cm}.\) What is the length of to the nearest tenth of a centimeter?
+983 You didn’t specify what we need to find, but assuming you want BC, here is my solution:
Using the Pythagorean Thereom, BD^2 = AB^2 - AD^2 = 13^2 - 5^2 = 12^2
BD = 12
Using the same method, BC^2 = DC^2 - BD^2 = 20^2 - 12^2
You take the square root of that and you have your answer.
+129852 And if you want to find AC ...
cos BDC = 12/20 = 3/5
cos BDA = 5/13
sin BDC = 4/5
sin BDA = 12/13
So cos (BDC + BDA) = cosBDCcosBDA - sinBDC sinBDA =
(3/5)(5/13) - (4/5)(12/13) =
15/65 - 48/65 =
-33/65
So....using the Law of Cosines
AC^2 = AD^2 + DC^2 - 2(AD * DC) cos ( BDC + BDA)
AC^2 = 5^2 + 20^2 - 2(5 * 20)(-33/65)
AC^2 = 425 + 200(33/65)
AC^2 = 6845 / 13
AC ≈ 22.946
