Angle of Depression Quiz: Air Traffic Control Angle of Depression

The answer is True. h = d×tanθ. When θ is constant, tanθ is a fixed multiplier. Doubling d multiplies the right side by 2, producing exactly twice the altitude. For example, if d = 10 km and θ = 18°, then h = 10×tan18° = 3.25 km. At d = 20 km, h = 20×tan18° = 6.49 km, which is exactly double. The relationship between altitude and horizontal distance is directly proportional at a fixed angle.

The answer is True. The horizontal at the tower and the horizontal at the aircraft are parallel lines. The line of sight acts as a transversal cutting both. By the alternate interior angles theorem from geometry, the angle formed below the tower horizontal equals the angle formed above the aircraft horizontal. The depression angle and elevation angle along the same line of sight are therefore always equal.

h = d×tanθ = 20.0×tan6° = 20.0×0.1051 = 2.10 km. Option B gives 1.75, requiring tan(θ) = 0.0875, corresponding to approximately 5°. Option C gives 2.50, requiring tan(θ) = 0.125, corresponding to approximately 7.1°. Option D gives 3.20, requiring tan(θ) = 0.16, corresponding to approximately 9.1°. Only 2.10 km satisfies the equation with θ = 6°.

The answer is False. h = d×tanθ. Doubling θ gives h_new = d×tan(2θ). Since tangent is nonlinear, tan(2θ) ≠ 2×tan(θ) in general. For example, tan(20°) = 0.364 but tan(40°) = 0.839 ≠ 2×0.364 = 0.728. Doubling the angle produces a different multiplier, so the altitude changes by a factor other than 2 for any angle except special cases.

h = 15.2×tan24° = 15.2×0.4452 = 6.77 km. Option B gives 7.07, which corresponds to 15.2×tan25° = 15.2×0.4663 = 7.09 km, the wrong angle. Option C gives 5.50, requiring tan(θ) = 0.362, corresponding to approximately 19.9°. Option D gives 8.20, requiring tan(θ) = 0.539, corresponding to approximately 28.3°. Only 6.77 km matches θ = 24°.

The answer is True. A calculator in radians mode interprets the input as radians. Entering 24 when the mode is radians evaluates tan(24 radians), not tan(24°). Since 24 radians is approximately 1,375°, the result is completely different from tan(24°) = 0.4452. Always verifying that the calculator mode matches the unit of the given angle is essential to avoid large calculation errors.

h = 13.75×tan19° = 13.75×0.3443 = 4.73 km. Option B gives 3.90, requiring tan(θ) = 0.2836, corresponding to approximately 15.8°. Option C gives 5.40, requiring tan(θ) = 0.3927, corresponding to approximately 21.4°. Option D gives 6.10, requiring tan(θ) = 0.4436, corresponding to approximately 23.9°. Only 4.73 km satisfies the equation with θ = 19°.

The altitude is the opposite side and the horizontal distance is the adjacent side, so tan(theta) = h divided by d, giving h = d times tan(theta). Option B gives d divided by tan(theta) = d times cot(theta), which is the formula for the adjacent side if the opposite were known, not altitude. Option C uses sine, which requires the slant distance as hypotenuse, not the horizontal distance. Option D gives d times tan(90 degrees minus theta) = d times cot(theta), the same incorrect result as option B.

In the right triangle formed by the tower, the aircraft, and the ground, the vertical altitude is opposite the angle and the horizontal separation is adjacent, confirming A and B. Tangent is defined consistently in both degree and radian systems as long as the input matches the mode, confirming C. h = d×tanθ follows directly from the definition of tangent, confirming D.

Option A is incorrect — it inverts the correct ratio. The right identity is tan(θ) = h/d, so writing tan(θ) = d/h is wrong. Option D is incorrect — h = d/tan(θ) does not give the slant height. It gives d×cot(θ), a different quantity entirely. Slant height requires h/sin(θ) or d/cos(θ). Options B and C are both correct statements and should not be selected.

h = 12.5×tan18° = 12.5×0.3249 = 4.06 km. Option B gives 3.20, requiring tan(θ) = 0.256, corresponding to approximately 14.4°. Option C gives 5.10, requiring tan(θ) = 0.408, corresponding to approximately 22.2°. Option D gives 2.85, requiring tan(θ) = 0.228, corresponding to approximately 12.8°. Only 4.06 km satisfies the equation with θ = 18°.

h = 11.3×tan17.5° = 11.3×0.3153 = 3.56 km. Option B gives 2.90, requiring tan(θ) = 0.257, corresponding to approximately 14.4°. Option C gives 4.20, requiring tan(θ) = 0.372, corresponding to approximately 20.4°. Option D gives 5.00, requiring tan(θ) = 0.442, corresponding to approximately 23.8°. Only 3.56 km satisfies the equation with θ = 17.5°.

h = 5.4×tan28° = 5.4×0.5317 = 2.87 km. Option B gives 2.67, requiring tan(θ) = 0.494, corresponding to approximately 26.3°. Option C gives 3.07, requiring tan(θ) = 0.569, corresponding to approximately 29.6°. Option D gives 3.50, requiring tan(θ) = 0.648, corresponding to approximately 33°. Only 2.87 km satisfies the equation with θ = 28°.

tanθ = h/d = 2.8/9.6 = 0.2917. θ = arctan(0.2917) = 16.26°, which rounds to 16°. At 15°, tan15° = 0.2679, giving h = 9.6×0.2679 = 2.57 km, not 2.8. At 17°, tan17° = 0.3057, giving h = 9.6×0.3057 = 2.93 km. At 18°, tan18° = 0.3249, giving h = 3.12 km. Only 16° satisfies the given values to the nearest degree.

h = 9.0×tan12° = 9.0×0.2126 = 1.91 km. Option B gives 1.60, requiring tan(θ) = 0.178, corresponding to approximately 10.1°. Option C gives 2.20, requiring tan(θ) = 0.244, corresponding to approximately 13.7°. Option D gives 2.55, requiring tan(θ) = 0.283, corresponding to approximately 15.8°. Only 1.91 km satisfies the equation with θ = 12°.

Option A provides horizontal distance d and angle θ, giving h = d×tan(θ) directly. Option D provides slant distance and angle θ, giving h = slant×sin(θ) since altitude is opposite θ and slant is the hypotenuse. Option B gives only d with no angle information, leaving h unknown. Option C gives only θ with no distance, leaving h unknown. Both B and C are individually insufficient to determine altitude.

The answer is True. h = d×tan0° = d×0 = 0. An angle of depression of 0° means the line of sight from the tower is perfectly horizontal. The aircraft is therefore at the same height as the observer in the tower. Since altitude in this context measures the vertical separation from the tower's eye level down to the aircraft, a horizontal line of sight produces zero vertical separation, confirming that the altitude equals 0.

h = 7.8×tan30° = 7.8×(1/√3) = 7.8×0.5774 = 4.50 km. tan30° = 1/√3 is a standard exact value. Option B gives 3.90, requiring tan(θ) = 0.5, corresponding to approximately 26.6°. Option C gives 5.20, requiring tan(θ) = 0.667, corresponding to approximately 33.7°. Option D gives 6.00, requiring tan(θ) = 0.769, corresponding to approximately 37.6°. Only 4.50 km satisfies θ = 30°.

h = d×tanθ increases as θ increases on the interval from 0° to 90°. tan35° = 0.7002 is the largest of the four values: tan10° = 0.1763, tan15° = 0.2679, tan25° = 0.4663, tan35° = 0.7002. Multiplying each by the same fixed d, the largest result comes from θ = 35°. A steeper angle of depression always corresponds to a greater altitude when the horizontal distance is held constant.

tan(θ) = h/d follows from opposite over adjacent, confirming A. cos(θ) = d/slant distance follows from adjacent over hypotenuse, confirming B. sin(θ) = h/slant distance follows from opposite over hypotenuse, confirming C. Option D states sin(θ) = d/slant distance, which is false — d is the adjacent side, not the opposite side, so this ratio gives cos(θ), not sin(θ).