Answer:
The length of EF is 5.29
Step-by-step explanation:
If BC = 12 then;
FB = 12 × sin65 = 10.876
BA = FB/(sin50) = 10.876/(sin50) = 14.2
BD = AB (Sides of isosceles triangle) = 14.2
∴ DF = BD + FB = 10.876 + 14.2 = 25.07
∠F C D = tan⁻¹(D F/F C) = tan⁻¹(25.07/(12×cos65)) = 78.6
∠F A D = tan⁻¹(D F/A F) = tan⁻¹(25.07/(14.2×cos50)) = 70
∴ ∠D = 180 - (78.6° + 70°) = 31.4°
AE = AD × sin31.4° = 2×AB×cos(70° - 50°) × sin31.4
∴ AE = 2×14.2×cos(20°) × sin31.4 = 13.92
EC = AC × cos78.6° = (AB×cos50 + BC×cos65)×cos78.6° = 2.815
Therefore, from cosine rule, a² = b² + c² - 2·b·c × cos(A)
Hence, we have;
EF² = EC² + FC² - 2 × EC × FC ×cos∠FCD (Cosine rule)
That is, EF² = 2.815² + (12×cos65)² - 2 × 2.815× (12×cos65) ×cos78.6°
EF² = 27.98
∴EF = √27.98 = 5.29.
