Asked by kud

AB = 8 cm, AC = 6 cm, AD = 7 cm, CD = 2.82 and CAB=50°.
(a) Find the length BC,
(BC)^2 = (AB)^2 +(AC)^2 -2(AB)(AC)cosA
(BC)^2 = 64 +36 - 96cos(50)
BC =sqroot100-96cos(50)
BC = 6.188
(b) angle ABC,

(c) angle CAD ,

(d) area of triangle ACD
Answered by Steve
sinB/6 = sin50/6.188

2.82^2 = g^2+7^2-2*6*7 cos(CAD)

area of ACD = 6*(7*sin(CAD))/2
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