Geometry
posted by stranger
Convex quadrilateral ABCD has sides AB=BC=21, CD=15 and AD=9. Given additionally that ∠ABC=60∘, what is the length of BD?

Reiny
Join the diagonals
Since B  60° and AB = BC = 21
ABC must be equilateral and AC = 21
Now in triangle ACD we have 3 sides, and by the cosine law I found angle D to be 120°
interesting!
using the sine law in triangle ACD
sin CAD/15 = sin60/21
sin CAD = .61858..
angle CAD = 38.3232... (I stored that)
angle BAD = 60 + 38.32... = 98.21..°
by the cosine law:
BD^2 = 21^2 + 9^2  2(21)(9)cos BAD
= 528
BD = √528 or appr 22.98 
Finalhero
Reiny in the last step u have done an error as 21^2 =441 + 9^2 = 81 = 522
and 2*21*9*cos(98) =  54.6
so 522  (54.6)=522+52.6=574.6
and square root of 574.6 = 23.97 ~ 24
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