+6
The side lengths of a cyclic quadrilateral ABCD are provided in the diagram. Find DB^2
+128406 Note that angle BAD is acute
Note that angle DCB is supplemental to angle BAD.....so cos DCB = - cosBAD
And by the Law of Cosines we have that
DB^2 = CB^2 + CD^2 - 2(CB)(CD)(-cos)(BAD)
DB^2 = BA^2 + AD^2 - 2(BA)(AD) cos(BAD) substituting we have that
DB^2 = 2^2 + 3^2 + 2(3)(2) cos BAD
DB^2 = 1^2 + 4^2 - 2(1)(4) cos BAD simplify
DB^2 = 13 - 12cosBAD ⇒ [ DB^2 - 13] / [-12 ] = cosBAD (1)
DB^2 = 17 - 8 cosBAD (2)
Sub (1) into (2) and we have that
DB^2 = 17 - 8 [ DB^2 - 13] / [-12]
DB^2 - 17 = -8[DB^2 - 13] / [-12]
DB^2 - 17 = (2/3)[DB^2 - 13 ]
DB^2 - 17 = (2/3)DB^2 - 26/3 subtract (2/3)DB^2 from both sides
(1/3)DB^2 - 17 = - 26/3 multiply through by 3
DB^2 - 51 = - 26 add 51 to both sides
DB^2 = 25 take the positive root
DB = 5
