Using Cachi's theorem for triangles ABC and BEG we find cos(a)=(AB²+BC²-AC²)/(2*AB*BC)=(BE²+BG²-EG²)/(2*BE*BG) then let AB=x. Then BC²=x²+16, BE²=x²+144, BG²=x²+576, so we get an equation with one unknown x. After simplification, we get an equivalent equation x⁴-72x²-10368=0, i.e. (x²+72)(x²-144)=0. Therefore, x=12, and from this, BG=12√5. Since the triangles ABG and CFG are similar and their similarity ratio is equal to CG,/BG=5/(3√5), the area of triangle CFG is (12*24/2)*(5/3√5)²=80.
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Using Cachi's theorem for triangles ABC and BEG we find cos(a)=(AB²+BC²-AC²)/(2*AB*BC)=(BE²+BG²-EG²)/(2*BE*BG) then let AB=x. Then BC²=x²+16, BE²=x²+144, BG²=x²+576, so we get an equation with one unknown x. After simplification, we get an equivalent equation x⁴-72x²-10368=0, i.e. (x²+72)(x²-144)=0. Therefore, x=12, and from this, BG=12√5. Since the triangles ABG and CFG are similar and their similarity ratio is equal to CG,/BG=5/(3√5), the area of triangle CFG is (12*24/2)*(5/3√5)²=80.
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φ = 30° → sin(3φ) = 1; ∆ ABG → AB = a; AG = AC + CE + EG = 4 + 8 + 12
BG = BF + GF → sin(CFB) = 1 = sin(BAG) ;CF = t; GBE = CBA = α; EBC = β; AGB = δ
BC = x; BE = y; BG = z → x = √(a^2 + 16); y = √(a^2 + 144); z = √(a^2 + 576)
(1/2)sin(α)ax = (1/3)(1/2)sin(α)xy → ax = yz/3 →
a√(a^2 + 16) = (1/3)√(a^2 + 144)√(a^2 + 576) → a = 12 →
sin(δ) = √5/5 = t/20 → t = 4√5 → FG = 8√5 → area ∆ CGF = 80
or: CF = t; AC = AE/3 → arctan(1) - arctan(1/3) = arctan(1/2) = tan(β) →
tan(α) = 1/3 → a = 12 → AGB = δ → sin(δ) = √5/5 = t/20 → t = 4√5 →
FG = 2t → area ∆ CGF = t^2 = 80 → x = 4√10 → y = 12√2 → z = 12√5
(4)^2(8)^2(12)^2={16+64144}=224ABCDEFG/180°=1.44ABCDEFG 1.2^2^2^2 1.1^1^1^2 1^2 (ABCDEFG ➖ 2ABCDEFG+1).
Solution by trigonometry: All lengths are in km and areas in km².