Solutions — Right-Triangle Trigonometry

1. Find the unknown side. Fluency

   • (a) 30°, hyp=10, find opp: opp = 10 × sin 30° = 10 × 1/2 = 5 (exact).
   • (b) 45°, hyp=8, find adj: adj = 8 × cos 45° = 8 × √2/2 = 4√2 ≈ 5.66.
   • (c) 60°, adj=6, find opp: tan 60° = opp/adj ⇒ opp = 6 × tan 60° = 6√3 ≈ 10.39.
   • (d) 25°, opp=5, find hyp: sin 25° = 5/hyp ⇒ hyp = 5/sin 25° = 5/0.4226 ≈ 11.83.

2. Find the angle. Fluency

   • (a) opp=3, hyp=6: sin θ = 3/6 = 0.5 ⇒ θ = sin²¹(0.5) = 30.0°.
   • (b) adj=5, hyp=10: cos θ = 5/10 = 0.5 ⇒ θ = cos²¹(0.5) = 60.0°.
   • (c) opp=4, adj=4: tan θ = 4/4 = 1 ⇒ θ = tan²¹(1) = 45.0°.
   • (d) opp=7, hyp=10: sin θ = 7/10 = 0.7 ⇒ θ = sin²¹(0.7) ≈ 44.4°.

3. Exact values. Fluency

   • (a) sin 30° + cos 60°: 1/2 + 1/2 = 1. (sin 30° = cos 60° since they are complementary angles.)
   • (b) tan 45° × cos 45°: 1 × √2/2 = √2/2.
   • (c) sin² 60° + cos² 60°: (√3/2)² + (1/2)² = 3/4 + 1/4 = 1. This is the Pythagorean identity: sin²θ + cos²θ = 1 for all θ.
   • (d) 30° angle, hyp=6, find opp exactly: opp = 6 × sin 30° = 6 × 1/2 = 3.

4. Solve the right triangle. Fluency

   • (a) A=50°, hyp=12: Remaining angle = 90°−50° = 40°. Opposite (a) = 12×sin50° ≈ 9.19 m. Adjacent (b) = 12×cos50° ≈ 7.71 m.
   • (b) A=35°, adj=8: Remaining angle = 90°−35° = 55°. Opposite = 8×tan35° ≈ 5.60 cm. Hypotenuse = 8/cos35° ≈ 9.76 cm.

5. Read from the diagram. Understanding

   • (a) Which side is opposite to 40°?: BC is opposite the 40° angle at A (it is the side that does not touch angle A).
   • (b) Trig equation: sin 40° = BC/8. (Opposite over hypotenuse.)
   • (c) Find BC: BC = 8 × sin 40° = 8 × 0.6428 ≈ 5.14 m.
   • (d) Find AB: AB = 8 × cos 40° = 8 × 0.7660 ≈ 6.13 m. (Or verify: AB² + BC² = 6.13²+5.14² ≈ 37.6+26.4 = 64 = 8² ✓)

6. Pythagoras and trigonometry combined. Understanding

   • (a) Hypotenuse: hyp = √(5²+12²) = √(25+144) = √169 = 13 cm.
   • (b) Two acute angles: θ₁ = tan²¹(5/12) ≈ 22.6°. θ₂ = tan²¹(12/5) ≈ 67.4°.
   • (c) Verify: 22.6° + 67.4° = 90.0° ✓.
   • (d) Verify using cos: adj = hyp × cos(θ₁) = 13 × cos 22.6° = 13 × 0.9231 ≈ 12.0 cm ✓.

7. Choose the correct ratio. Understanding

   • (a) θ, opposite, hypotenuse: Sine: sin θ = opposite/hypotenuse. (SOH)
   • (b) θ, adjacent, hypotenuse: Cosine: cos θ = adjacent/hypotenuse. (CAH)
   • (c) θ, opposite, adjacent: Tangent: tan θ = opposite/adjacent. (TOA)
   • (d) 55°, adj=9, find hyp: cos 55° = 9/hyp ⇒ hyp = 9/cos55° = 9/0.5736 ≈ 15.69 m.

8. Perimeter of a right triangle. Understanding

   • (a) Opposite to 38°: opp = 20 × sin 38° = 20 × 0.6157 ≈ 12.31 cm.
   • (b) Adjacent to 38°: adj = 20 × cos 38° = 20 × 0.7880 ≈ 15.76 cm.
   • (c) Perimeter: P = 12.31 + 15.76 + 20 = 48.07 cm.
   • (d) Area: A = ½ × base × height = ½ × 15.76 × 12.31 ≈ 96.96 cm².

9. Ladder problem. Problem Solving

   • (a) Height on wall: h = 5 × sin 70° = 5 × 0.9397 ≈ 4.70 m.
   • (b) Base distance from wall: d = 5 × cos 70° = 5 × 0.3420 ≈ 1.71 m.
   • (c) Height at 75°: h = 5 × sin 75° = 5 × 0.9659 ≈ 4.83 m.
   • (d) Angle for 4 m height: sin θ = 4/5 = 0.8 ⇒ θ = sin²¹(0.8) ≈ 53.1°.

10. Sun and shadow. Problem Solving

    • (a) Diagram: Horizontal ground with the shadow (3.2 m) as the adjacent side, vertical post (h) as the opposite side, and the sun’s rays as the hypotenuse, meeting the ground at 35°.
    • (b) Equation: tan 35° = h / 3.2. (Opposite = post height h; adjacent = shadow length 3.2 m.)
    • (c) Height of post: h = 3.2 × tan 35° = 3.2 × 0.7002 ≈ 2.24 m.
    • (d) Shadow equals post height: shadow = h ⇒ tan θ = h/h = 1 ⇒ θ = 45°. When the sun is at 45°, every vertical object casts a shadow equal to its own height.