riangular flower garden ABC in which AB = 4m, BC = 5and ∠BCA =30. Point D lies on AC such that BD = 4 m and ∠BDC is obtuse.
Find:
(a) ∠BDC
sinD/a = sinC/b
sinD/5 = sin30/4
sinD = 5/8
This is a part I stuck to figure
(b) the length of AD
ADsin/4 =
(c) the length of DC
Sin30/5 = DC/4
(d) the area of the flower garden ABC
1/2(4)(5)sin(30)
This is about the 3rd time I have helped you with this question, the last time was this afternoon.
I thought we had everything cleared up, you are using mostly the sine law.
I can't find my previous posts, but I have it here before me on a sheet of paper.
a) Picking it from
sin D/5 = sin 30°/4
sinD = 5(1/2) / 4 = 5/8
D = 38.68° or 180-38.68 or 141.32° by the CAST rule
The question stated the angle BDC was obtuse, so we have to go with
∠BDC = 141.32°
b) Look at triangle ABD, it is isosceles, ∠BDA = ∠BAD = 38.68°
so ∠ABD = 102.64°
Then AD/sin102.64 = 4/sin38.68
this gives me AD = 6.245
c) the length of DC
Again, the sine law:
We need ∠CBD
∠CBD + 30 + 102.64 = 180
∠CBD = 47.36°
CD/sin47.36 = 4/sin30
CD = 4sin47.36/sin30 = ....
d) area of ABC = (1/2)(5)(4)sin (102.64°+ 47.36°) = ......