Navigation & Flight Planning

Correct

The answer is B: 90 NM

A degree of longitude spans about 60 × cos(latitude) NM. At 60°N, cos 60° = 0.5, so each degree is about 30 NM and 3° ≈ 90 NM. 180 NM wrongly treats longitude degrees like latitude degrees.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is C: 6,076 feet

A nautical mile is about 6,076 ft, roughly 15 percent longer than a statute mile (a ratio of about 1.15). A knot is one nautical mile per hour.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is A: TH 080°, GS 118 kt

The wind is a direct left crosswind. WCA = asin(20 × sin 270° ÷ 120) ≈ −9.6°, so TH ≈ 080°. GS = 120 × cos 9.6° − 20 × cos 270° ≈ 118 kt. Correcting right (100°) applies the wind backward; a pure crosswind doesn't subtract its full speed from TAS.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is C: TH 277°, GS 83 kt

The wind is 30° to the right of the nose. WCA = asin(25 × sin 30° ÷ 105) ≈ +6.8°, so TH ≈ 277°. GS = 105 × cos 6.8° − 25 × cos 30° ≈ 104.3 − 21.7 ≈ 83 kt. Correcting left (263°) or treating the headwind as a tailwind (127 kt) are the common errors.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is B: TH 056°, GS 129 kt

The wind is from the right rear (135° right of the course). WCA = asin(30 × sin 135° ÷ 110) ≈ +11.1°, so TH ≈ 056°. GS = 110 × cos 11.1° − 30 × cos 135° ≈ 108 + 21 ≈ 129 kt. It is a quartering tailwind, so GS exceeds TAS.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is B: TH 312°, GS 82 kt

The wind is 50° left of the course. WCA = asin(18 × sin(−50°) ÷ 95) ≈ −8.3°, so TH ≈ 312°. GS = 95 × cos 8.3° − 18 × cos 50° ≈ 94.0 − 11.6 ≈ 82 kt. Correcting right (328°) applies the wind backward.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is A: 7° left; TH 118°

The wind is 65° left of the course. WCA = asin(20 × sin(−65°) ÷ 140) ≈ −7.4°, so TH ≈ 118°. Correcting right (132°) would add to the drift instead of countering it; the groundspeed works out to about 130 kt.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is C: 17-kt headwind and 10-kt crosswind from the right

The wind is 30° right of the course. Headwind = 20 × cos 30° ≈ 17 kt; crosswind = 20 × sin 30° = 10 kt, from the right because 030° is clockwise of 360°.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is B: 106 kt

WCA = asin(30 × sin 40° ÷ 130) ≈ 8.5°. GS = 130 × cos 8.5° − 30 × cos 40° ≈ 128.6 − 23.0 ≈ 106 kt, because a wind 40° off the nose gives a headwind component of only about 23 kt. 153 kt adds that component as if it were a tailwind, and 100 kt subtracts the whole 30 kt of wind speed.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is A: TH 090°, GS 135 kt

A wind from 270° on a 090° course is a direct tailwind: no crosswind, so no wind correction (TH = TC = 090°), and GS = 110 + 25 = 135 kt. Wind direction is where the wind comes FROM, so this wind pushes the airplane east.

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is C: Heading left of course; groundspeed greater than TAS

A wind from 330° on a 090° course comes from the left rear. The left crosswind component requires a left correction (TH < 090°), and the tailwind component makes GS greater than TAS. (TAS 120, 20 kt gives TH ≈ 082° and GS ≈ 129 kt.)

Source: PHAK Ch. 16 · ACS PA.VI.A

Correct

The answer is B: about 8 minutes longer

Into the wind: 100 NM ÷ 75 kt = 80 min. With the wind: 100 NM ÷ 125 kt = 48 min. Total = 128 min, compared with 120 min in calm air. You spend more time in the headwind than in the tailwind, so wind never cancels out on a round trip.

Source: PHAK Ch. 16 · ACS PA.VI.A