Help on this trigonometry problem...?

Question

In a quadrilateral ABCD, AB = 3.2 cm, BC = 5.1 cm, the diagonal BD = 7.5 cm and angle CBD = 34.4°. The area of ABD is 11.62 cm2 and angle ABD is obtuse. Calculate

a) the area of BCD,

b) the size of the obtuse angle ABD.

Answer 1

S= ab/2* sin v

here a nad b - sides of the triangle, v- angle between them

U need it to calculate everything

1) triangle BCD (CBD- is obtuse)

  S= BD*BC/2* sin(CBD)= 7.5*5.1/2*sin34.4=19.12*sin34.4

2) (ABD- is obtuse)

sin(pi- ABD)=2S(ABD)/( AB*BD)= 2*11.62/ (3.2* 7.5)= 0.968

Angle ABD= pi- arcsin0.968=

Answer 2

CD={5.1^2 + 7.5^2 - 2(5.1)(7.5)Cos 34.4} ^ 1/2,CD=4.375

Area of triangle={p(p-a)(p-b)(p-c)} ^ 1/2 where p=(a+b+c)/2

area of BCD={8.4875x(8.4875-7.5)(8.4875-5.1)(8.4... ^ 1/2=10.8 cm^2

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Use the area formulla to find ABD