Two converging aircraft on radials diverging by 15 degrees: at what point will separation be lost?

Two converging aircraft on radials diverging by 15 degrees: at what point will separation be lost?

• A. 9 miles
• B. 17 miles
• C. 25 miles
• D. 40 miles

Answer: (B)

When two aircraft sit on radials that diverge by a small angle, their lateral separation is tied to how far they are from the point where the radials cross. If both aircraft are about the same distance r from that intersection, the straight-line separation between them forms a triangle with two sides r and included angle θ, so the base (the separation) is s = 2 r sin(θ/2). Here θ is 15 degrees, so θ/2 is 7.5 degrees and sin(7.5°) ≈ 0.1305. That makes s ≈ 2 r × 0.1305 ≈ 0.261 r. Separation is considered lost when this lateral separation s equals the required minimum separation (the standard en-route minimum is commonly about 5 miles). Solving for r gives r ≈ 5 / 0.261 ≈ 19 miles. Of the given options, 17 miles is the closest practical distance, which is why that choice is the best answer.

When two aircraft sit on radials that diverge by a small angle, their lateral separation is tied to how far they are from the point where the radials cross. If both aircraft are about the same distance r from that intersection, the straight-line separation between them forms a triangle with two sides r and included angle θ, so the base (the separation) is s = 2 r sin(θ/2). Here θ is 15 degrees, so θ/2 is 7.5 degrees and sin(7.5°) ≈ 0.1305. That makes s ≈ 2 r × 0.1305 ≈ 0.261 r.

Separation is considered lost when this lateral separation s equals the required minimum separation (the standard en-route minimum is commonly about 5 miles). Solving for r gives r ≈ 5 / 0.261 ≈ 19 miles. Of the given options, 17 miles is the closest practical distance, which is why that choice is the best answer.