Angle A of triangle ABC will be ...
∠A = θ₂ -θ₃ = 77° -26° = 51°
Angle B of triangle ABC will be ...
∠B = 180° -θ₂ -θ₁ = 180° -77° -44° = 59°
Then angle C will be ...
∠C = 180° -∠A -∠B = 180° -51° -59° = 70°
By the Law of Sines,
a/sin(∠A) = c/sin(∠C)
a = c×sin(∠A)/sin(∠C)
a = (57 yd)×sin(51°)/sin(70°) ≈ 47.14 yd
The distance from B to C is 47.14 yd.
