Statue Height Quiz: Statue Height From Two Angles

The answer is False. Knowing h and θ_b determines only the horizontal distance d = h/tanθ_b. To find H, the angle to the top θ_t is also needed, since H = h − d×tanθ_t. Without θ_t, the height of the statue's top above the ground cannot be determined and H remains unknown. At least three independent pieces of information are required to solve for H uniquely.

Option A is valid: d = h/tanθ_b is derived from tanθ_b = h/d, then substituting into H = h − d×tanθ_t gives the correct statue height. Option C is also valid: substituting d = h/tanθ_b into H = d(tanθ_b − tanθ_t) gives the same result through an equivalent path. Option B is wrong because H = h×(tanθ_b − tanθ_t) is not a valid formula — the correct substitution gives H = (h/tanθ_b)×(tanθ_b − tanθ_t). Option D is wrong because d×tanθ_t − h gives a negative value in this geometry.

d = h/tanθ_b = 30/tan40° = 30/0.8391 = 35.75 m. H = h − d×tanθ_t = 30 − 35.75×tan10° = 30 − 35.75×0.1763 = 30 − 6.30 = 23.70 m. Option B gives 20.50, requiring d×tanθ_t = 9.50. Option C gives 26.40, requiring d×tanθ_t = 3.60. Option D gives 18.90, requiring d×tanθ_t = 11.10. None match the computed value of 6.30.

The vertical drop from eye level to the base is d×tanθ_b and to the top is d×tanθ_t. Their difference gives H = d(tanθ_b − tanθ_t), confirming A and C. When h is known, H = h − d×tanθ_t because h is the full drop to the ground and d×tanθ_t is the drop to the statue top, confirming B. Option D, H = h×(tanθ_t/tanθ_b), has no valid geometric derivation in this setup.

The answer is True. The horizontal at the observer's level and the horizontal at the lower point are parallel. The line of sight acts as a transversal cutting these two parallel lines. By the alternate interior angles theorem, the angle of depression measured downward from the observer's horizontal equals the angle of elevation measured upward from the lower horizontal. The two angles are always equal along the same line of sight.

H = 36×(tan32° − tan6°) = 36×(0.6249 − 0.1051) = 36×0.5198 = 18.71 m. Option B requires the tangent difference to equal 0.622. Option C requires 0.425. Option D requires 0.697. None of these match the computed difference of 0.5198 for angles 32° and 6°.

The answer is False. H = d(tanθ_b − tanθ_t). Doubling the angles gives H_new = d(tan(2θ_b) − tan(2θ_t)). Since tangent is nonlinear, tan(2θ) ≠ 2×tan(θ) in general. For example, tan(30°) = 0.577 but tan(60°) = 1.732 ≠ 2×0.577 = 1.154. The new difference of tangents is not twice the original, so H does not double.

Option A gives d and both angles, so H = d(tanθ_b − tanθ_t) is directly computable. Option B gives h and both angles, so d = h/tanθ_b can be found first, then H = h − d×tanθ_t. Option C gives only h and θ_b, which determines d but not the top angle, leaving H unknown. Option D gives only d and θ_t, which determines the drop to the top but not to the base.

h = d×tanθ_b = 50×tan28° = 50×0.5317 = 26.59 m. Option B gives 22.40, requiring tan(θ) = 0.448, corresponding to approximately 24.2°, not 28°. Option C gives 30.10, requiring tan(θ) = 0.602, corresponding to approximately 31°. Option D gives 18.75, requiring tan(θ) = 0.375, corresponding to approximately 20.6°. Only 26.59 m satisfies the equation.

H = d(tanθ_b − tanθ_t) = 60×(tan35° − tan12°) = 60×(0.7002 − 0.2126) = 60×0.4876 = 29.26 m. Option B requires the tangent difference to equal 0.5710. Option C requires 0.4043. Option D requires 0.6543. None of these match the computed difference of 0.4876 for angles 35° and 12°.

The answer is True. H = d(tanθ_b − tanθ_t). Since d and θ_t are fixed, tanθ_t is constant. As θ_b increases, tanθ_b increases, making the difference (tanθ_b − tanθ_t) larger. Multiplying by the fixed positive d gives a larger H. The relationship is monotonically increasing with θ_b, meaning every increase in θ_b produces a strictly greater statue height.

The height of the statue top above the ground = h − H = 26 − 18 = 8 m. tanθ_t = (h − H)/d = 8/40 = 0.2. θ_t = arctan(0.2) = 11.31°, which rounds to 11°. Option B gives 14°, where tan14° = 0.2493, giving h − H = 9.97 m, not 8 m. Option C gives 8°, where tan8° = 0.1405. Option D gives 17°, where tan17° = 0.3057. Only 11° satisfies the given values.

tanθ_b = h/d = 24/55 = 0.4364. θ_b = arctan(0.4364) = 23.57°, which rounds to 24°. Option B gives 26°, where tan26° = 0.4877, giving h = 55×0.4877 = 26.8 m, not 24 m. Option C gives 22°, where tan22° = 0.4040, giving h = 22.2 m. Option D gives 29°, where tan29° = 0.5543, giving h = 30.5 m. Only 24° satisfies the given values.

Rearranging H = d(tanθ_b − tanθ_t) gives d = H/(tanθ_b − tanθ_t) = 18/(tan30° − tan8°) = 18/(0.5774 − 0.1405) = 18/0.4369 = 41.21 m. Option B gives d = 36.21, requiring the denominator to equal 0.497. Option C gives 46.21, requiring 0.390. Option D gives 51.21, requiring 0.352. None match the computed denominator of 0.4369.

The answer is True. Although angles of depression point downward, the right triangle formed by the horizontal, the vertical drop, and the line of sight uses positive acute angles. The tangent of the depression angle equals the vertical drop divided by the horizontal distance, with all values positive. This is why the formula H = d(tanθ_b − tanθ_t) works directly without any sign adjustments.

The base is further below the observer than the top, so the depression angle to the base is always steeper: θ_b > θ_t, confirming A. Equal angles give tanθ_b − tanθ_t = 0, so H = 0, confirming B. Option C is false — θ_t = 0° means the top is at eye level, giving H = h only if the geometry specifically places the top at observer height, which is not automatic. H = d(tanθ_b − tanθ_t) scales directly with d when angles are fixed, confirming D.

d = h/tanθ_b = 20/tan35° = 20/0.7002 = 28.56 m. H = h − d×tanθ_t = 20 − 28.56×tan18° = 20 − 28.56×0.3249 = 20 − 9.28 = 10.72 m. Option B gives 12.80, which would require d×tanθ_t = 7.20, meaning d = 22.16 m. Option C gives 8.50, requiring d×tanθ_t = 11.50. Option D gives 14.20, requiring d×tanθ_t = 5.80. None match the computed values.

H = 42×(tan33° − tan7°) = 42×(0.6494 − 0.1228) = 42×0.5266 = 22.12 m. Option B requires the tangent difference to equal 0.598, option C requires 0.455, and option D requires 0.670. None of these match the computed difference of 0.5266 for these angles.

H = d(tanθ_b − tanθ_t) = 30×(tan25° − tan9°) = 30×(0.4663 − 0.1584) = 30×0.3079 = 9.24 m. Option B requires the difference of tangents to equal 0.4167, which does not match any standard angle pair at d = 30. Option C requires 0.26 and option D requires 0.37. Neither matches the computed difference of 0.3079.

Option A is a mistake because tan is not distributive over subtraction. Option B is a mistake because θ_b must always be greater than θ_t when the observer is above the statue top. Option C is a mistake because ignoring the tan ratio leads to incorrect vertical drop calculations. Option D is not a mistake — H = d times (tan(θ_b) minus tan(θ_t)) is the correct formula for statue height and should be used, not avoided.