Aviation Math - REVISION - CHAP4 - Lesson 4.6

If the True Heading is 145° and the Magnetic Heading is 137°, then:

• Variation equals 8 degrees

• Deviation equals 8 degrees

• Inclination equals 8 degrees

• None of the above

An aircraft is flying with a True Heading of 210° and a Magnetic Heading of 202°. What is the magnetic variation?

• 8°

• 18°

• Deviation equals 8 degrees

• None of the above

If an aircraft’s Magnetic Heading is 095°, and its True Heading is 100°, what is the variation?

• 15°

• 5°

• Deviation equals 5 degrees

• None of the above

If the Magnetic Heading of an aircraft is 280° and the True Heading is 270°, what is the magnetic variation?

• 1°

• 10°

• Deviation equals 10 degrees

• None of the above

Using the Bearing Shortcut Rule: θ = tan⁻¹((AB sin θ1 + BC sin θ2) / (AB cos θ1 + BC cos θ2)) Where θ is the total bearing angle, AB is the distance from A to B, BC is the distance from B to C, θ1 is the angle from A to B, θ2 is the angle from B to C.

An aircraft departs from Point X on a bearing of 045° towards Point Y, which is 250 km away. At Point Y, it changes course to Point Z, which is 180 km away, on a new bearing of 080°. Find the total bearing from X to Z.

• 59.6°

• 52.3°

• 75.0°

• 68.9°

Using the Bearing Shortcut Rule and the formula for difference in bearing: α = θ - θ1 Where α is the angle difference in bearing, θ is the total bearing angle, θ1 is the angle from A to B.

An aircraft departs from Point X on a bearing of 045° towards Point Y, which is 250 km away. At Point Y, it changes course to Point Z, which is 180 km away, on a new bearing of 080°. Find: the difference in bearing between the original course (X to Y) and the total bearing from X to Z.

• 14.6°

• 25.0°

• 10.2°

• 35.8°

Using the Bearing Shortcut Rule: θ = tan⁻¹((AB sin θ1 + BC sin θ2) / (AB cos θ1 + BC cos θ2))

Where θ is the total bearing angle, AB is the distance from A to B, BC is the distance from B to C, θ1 is the angle from A to B, θ2 is the angle from B to C.

A helicopter departs from Point A on a bearing of 30° towards Point B (400 km away). At B, it turns towards Point C (275 km away) on a new bearing of 40°. Find the total bearing from A to C. (Give your answer in degrees.)

• 34.1°

• 28.4°

• 45.2°

• 38.9°

A ship leaves Point M on a bearing of 20° towards Point N, which is 320 km away. Upon reaching N, it changes course to Point P (190 km away) on a new bearing of 60°. What is the total bearing from M to P?

• 44.1°

• 83.5°

• 34.7°

• 10.4°

An aircraft sets off from A on a bearing of 070° towards B (800 km) away. At B it changes course towards C (4000 km) away on 080°. What is the difference in bearing between the bearing from A to C and the one from A to B?

• 8.3°

• 5.2°

• 15°

• 148.3°

MM Bearing: An aircraft sets off from A on a bearing of 045° towards B (300 km) away. At B it changes course towards C (300 km) away on 060°. Find the bearing from A to C. Given: (Sin 45 = 0.707, Cos 45 = 0.707, Sin 60 = 0.866, Cos 60 = 0.5, Inverse tan (1.3032) = 52.5, Inverse tan (0.76732) = 37.5, inverse tan (0.89) = 41.6, sin 52.5 = 0.7933, cos 52.5 = 0.60876, tan 52.5 = 1.3032, sin 37.5 = 0.6087, cos 37.5 = 0.7933).

• A. 37.5 degrees

• B. 45 degrees

• C. 52.5 degrees

• D. None of the above