Trigonometry

The vertical difference between pole tops is 5m (11m - 6m), and the horizontal distance between their feet is 12m. This forms a right triangle, so the distance between tops equals sqrt(5^2 + 12^2) = sqrt(169) = 13m by Pythagorean theorem. Options A (12m) and C (14m) are incorrect calculations, while D (15m) doesn't match the computed result.

The equation x² + 2xy cotθ - y² = 0 represents two lines through origin. For homogeneous equation Ax² + 2Hxy + By² = 0, the angle φ between lines satisfies tanφ = 2√(H² - AB)/|A+B|. Here A=1, H=cotθ, B=-1. tanφ = 2√(cot²θ + 1)/|1-1| = 2√(cosec²θ)/0 = 2cosecθ/0 → ∞, so φ = 90°. The lines are perpendicular.

Using tan(30°) = 20/d1 and tan(60°) = 20/d2 where d1 and d2 are distances from each man to the flagstaff. Distance between men = d1 + d2. d1 = 20/0.577 = 34.64m, d2 = 20/1.732 = 11.55m. Total = 46.19m ≈ 46.18m (rounding).

Since tan(2A) = cot(A - 21°) and cot θ = tan(90° - θ), we have tan(2A) = tan(90° - (A - 21°)) = tan(111° - A). Therefore 2A = 111° - A, giving 3A = 111° and A = 37°. We verify 2A = 74° is acute, satisfying the condition. Options A, B, C result from algebraic errors in solving the equation.

Let ground point be P. For lower plane at 500m with 45° elevation: tan 45° = 500/d, giving d = 500m (horizontal distance). For upper plane with 60° elevation: tan 60° = h/500, giving h = 500√3 ≈ 866m. Distance between planes = 866 - 500 = 366m. Answer B is correct.

Write cot A + cosec A = cos A/sin A + 1/sin A = (cos A + 1)/sin A = 3. Square: (1 + cos A)²/sin²A = 9. Using sin²A = 1 - cos²A = (1-cos A)(1+cos A), we get (1+cos A)/(1-cos A) = 9. Solving gives cos A = 4/5. Option A (1), B (1/2), C (3/4) don't satisfy the equation.

Let satellite height be h. From station 1 (30°): tan30° = h/d₁, from station 2 (60°): tan60° = h/d₂. Distance between stations = d₁ - d₂ = 4000 km. So h/tan30° - h/tan60° = 4000. h(√3 - 1/√3) = 4000. h(3/√3 - 1/√3) = 4000. h(2/√3) = 4000, h = 2000√3 ≈ 3464 km. Option C is correct.

For a flag staff with angle of elevation 30° from a point 200 meters away, use the tangent formula: tan(30°) = height/200. Since tan(30°) = 1/√3, the height = 200 × (1/√3) = 200/√3 ≈ 115.47 meters. Option A (115.47 meter) is correct. The other options don't match this calculation.

Using the tangent function, let h be the height. From point A: tan30° = h/d, so d = h√3. From point B: tan60° = h/(d-20), so d-20 = h/√3. Substituting d: h√3 - 20 = h/√3. Solving: h√3 - h/√3 = 20, which gives h(3-1)/√3 = 20, so 2h/√3 = 20, hence h = 10√3 m.

The height difference between poles = 11 - 6 = 5 m. Horizontal distance between feet = 12 m. These form a right triangle where the distance between tops is the hypotenuse. Using Pythagoras: Distance = √(5² + 12²) = √(25 + 144) = √169 = 13 m.

Using complementary angle relationship: tan θ = cot(90° - θ). So tan 2A = cot (A - 18°) implies 2A = 90° - (A - 18°) = 108° - A. Therefore 3A = 108°, giving A = 36°.

Let the height be h and initial distance be 80 m. If initial angle is θ, then tan θ = h/80. When distance becomes 30 m (80-50), angle becomes 2θ, so tan 2θ = h/30. Using double angle formula tan 2θ = 2tan θ/(1 - tan²θ) and substituting: h/30 = 2(h/80)/(1 - h²/6400). Solving gives h = 40 m. This requires trigonometric identities.

Height difference = 80 - 63.5 = 16.5m. Since line joining tops makes 45° with horizontal, tan 45° = opposite/adjacent = 16.5/distance. tan 45° = 1, so distance = 16.5m.

Let height = 500m. For angle 45°, distance = 500/tan45° = 500m. For angle 60°, distance = 500/tan60° = 500/√3 ≈ 288.7m. Distance between planes = 500 - 288.7 = 211.3m. But this is horizontal distance. Using √3-1: 500(√3-1) = 500(1.732-1) = 366m.

Using the complementary angle relationship: if tan(A) = cot(B), then A + B = 90°. Substituting: (2θ + 45°) + 3θ = 90°, so 5θ = 45°, giving θ = 9°. Verify: tan(2×9° + 45°) = tan(63°) and cot(3×9°) = cot(27°), and since 63° + 27° = 90°, they are equal.