Question

In triangle ABC, angle A=50°, angle C=65°. Point F is on AC such that, BF is perpendicular to AC. D is a point on BF (extended) such that AD=AB. E is a point CD such that, AE is perpendicular to CD. If BC=12, what is the length of EF?

Answer 1

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.