+0  
 
+2
187
1
avatar+6 

 

The side lengths of a cyclic quadrilateral ABCD  are provided in the diagram. Find DB^2
 

 May 27, 2019
 #1
avatar+109560 
+1

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

 

 

cool cool cool

 May 27, 2019

22 Online Users

avatar
avatar
avatar
avatar