FInd AC and angle BCD

Question

This problem refers to the figure below, which shows a diagram for a bike frame.

Given BC = 49 cm, BD = 63 cm, CD = 76 cm, ∠ABC = 52°, and ∠ACB = 67°, find the following. (Round your answers to the nearest whole number.)

(a) the length of the chainstay, AC

AC =  _____cm

(b)    ∠BCD

∠BCD = _____ °

Answer 1

m∠A = 61º

AC / sin 52º = 49 / sin 61º

Answer 2

I'm using an "oblique" formula:

β=acos((a^2+c^2-b^2)/(2ac)

In this case: acosβ(49^2+76^2-63^2)/(2*49*76)

β=59.6°; in this case >bcd.

As for side ac; asa was given.

α=52°

β=61°

γ=67°

bc=side a=49

ab= " c

ac= " b

a/sinα=b/sinβ=c/sinγ

49/.788=62.182=b/.874; b=54.3855......

PS

Would of been easier if the display was labeled correctly.