+1537 Find the area of triangle ABC.
Given: angle A = 2 * angle C, AB = 36, BC = 60.
This is a really hard problem!
+129852 A
36
B 60 C
Law of Sines :
sin A / 60 = sin C / 36
sin A / sin C = 60 / 36
sin (2C) / sin C = 5 / 3
[ 2sin C cos C] / sin C = 5/3
2cosC = 5/3 divide both sides by 2
cosC = 5/6
arccos (5/6) = C ≈ 33.56°
A = 2C ≈ 67.12°
Angle B ≈ 180 - 67.12 - 33.56 ≈ 79.32°
approx area = (1/2)(BC)(BA) sin (B) = (1/2) (60)(36)sin(79.32) ≈ 1061.3 units^2
