Introduction to Trigonometry — Solutions

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1. Label triangle sides

   1. PQ, QR, PR; θ at P: Hypotenuse = PR (opposite right angle at Q). Opposite = QR (across from θ at P). Adjacent = PQ (between θ and right angle).
   2. AB, BC, AC; θ at A: Hypotenuse = AC (opposite right angle at B). Opposite = BC (across from θ at A). Adjacent = AB (between θ and right angle).
   3. DE, EF, DF; θ at D: Hypotenuse = DF (opposite right angle at E). Opposite = EF (across from θ at D). Adjacent = DE (between θ and right angle).
   4. Same triangle, θ at F: Hypotenuse = DF (unchanged). Opposite = DE (now across from θ at F). Adjacent = EF (now between θ and right angle). The opposite and adjacent swap.
   5. JK, KL, JL; θ at J: The right angle is at K, so the hypotenuse is JL (opposite K).
   6. XYZ; right angle at Y, θ at X: Opposite = YZ (across from X). Adjacent = XY (between X and right angle Y).
   7. RST; right angle at S, θ = angle R: Opposite = ST (across from R). Adjacent = RS (between R and right angle S).
   8. MNO; right angle at N, θ at O: Hypotenuse = MO (opposite right angle at N). Opposite = MN (across from θ at O). Adjacent = NO (between θ at O and right angle N).

2. Identify which ratio to use

   1. Opposite and hypotenuse: sine — sin θ = O/H
   2. Adjacent and hypotenuse: cosine — cos θ = A/H
   3. Opposite and adjacent: tangent — tan θ = O/A
   4. Find hypotenuse, know opposite: sine — sin θ = O/H, rearrange to H = O/sin θ
   5. Find adjacent, know hypotenuse: cosine — cos θ = A/H, rearrange to A = H×cos θ
   6. Find opposite, know adjacent: tangent — tan θ = O/A, rearrange to O = A×tan θ
   7. Adjacent = 7, hypotenuse = 25: cosine — cos θ = 7/25
   8. Opposite = 9, adjacent = 40: tangent — tan θ = 9/40

3. Set up fractions without solving

   1. sin(42°) = 8/12: sin(42°) = 8/12  (or equivalently 2/3)
   2. cos(28°) = 15/17: cos(28°) = 15/17
   3. tan(55°) = 11/9: tan(55°) = 11/9
   4. sin(63°) = x/20: sin(63°) = x/20
   5. cos(37°) = x/14: cos(37°) = x/14
   6. tan(48°) = x/6: tan(48°) = x/6

4. Choosing the correct ratio and reasoning

   1. Keanu’s error: Keanu used cos but the given sides are opposite and hypotenuse, which belong to sine. Correct equation: sin(50°) = 13/17.
   2. Triangle ABC, angle B = 32°:
      1. AB is the adjacent side (between angle B and the right angle at A). BC is the hypotenuse (opposite the right angle at A).
      2. No, Priya’s equation is wrong. AB is adjacent and BC is hypotenuse, so the correct ratio is cosine: cos(32°) = AB/BC = 9/17.
   3. All three ratios: sin(36°) = 7.3/12.4  •  cos(36°) = 10.0/12.4  •  tan(36°) = 7.3/10.0
   4. Why sin ≤ 1: sin θ = O/H. In any right triangle, the opposite side (O) is always shorter than the hypotenuse (H) because the hypotenuse is the longest side. Therefore O/H < 1, so sin θ can never exceed 1.

5. Match diagrams to equations

   1. Ladder problem:
      1. sin θ = 4/5  •  cos θ = 3/5  •  tan θ = 4/3
      2. Use sin θ = opposite/hypotenuse = wall/ladder. This gives sin θ = 4/5.
   2. Ramp problem:
      1. The rise of 1.5 m is the opposite side and the horizontal distance of 6 m is the adjacent side.
      2. tan θ = opposite/adjacent = 1.5/6
   3. Three students: Yes, all three equations are correct. Checking: sin α = O/H = 5/13 ✓  •  cos α = A/H = 12/13 ✓  •  tan α = O/A = 5/12 ✓

6. Mixed trig ratio challenge

   1. sin α, cos α, tan α: sin α = BC/AC = 24/26 = 12/13. cos α = AB/AC = 10/26 = 5/13. tan α = BC/AB = 24/10 = 12/5.
   2. sin γ, cos γ, tan γ: sin γ = AB/AC = 10/26 = 5/13. cos γ = BC/AC = 24/26 = 12/13. tan γ = AB/BC = 10/24 = 5/12.
   3. Why sin α = cos γ: The two acute angles of a right triangle are complementary (they sum to 90°). The side opposite α is the same side adjacent to γ, and vice versa. So O/H for α equals A/H for γ — sin and cos swap for complementary angles.

7. Setting up from a description

   1. All three ratios for θ: Hyp = √(6²+4²) = √52 = 2√13. tan θ = 4/6 = 2/3. sin θ = 4/(2√13) = 2/√13. cos θ = 6/(2√13) = 3/√13.
   2. Equation to find rafter if only rise and angle known: sin θ = rise/rafter → rafter = rise/sin θ (do not evaluate)
   3. Why tan avoids hypotenuse: tan θ = rise/run uses only the two known measurements (rise and horizontal run). The hypotenuse is the rafter length — the unknown they want to find. Using tan avoids needing to know a value you haven’t calculated yet.

8. Decide and justify

   1. θ=58°, hyp=15, find opp: sin(58°) = opp/15 → opp = 15×sin(58°). Use sine (opp and hyp known).
   2. θ=31°, opp=9, find adj: tan(31°) = 9/adj → adj = 9/tan(31°). Use tangent (opp and adj; no hyp needed).
   3. θ=44°, adj=11, find hyp: cos(44°) = 11/hyp → hyp = 11/cos(44°). Use cosine (adj and hyp).
   4. Know all 3 sides, find an angle: All three ratios (sin, cos, tan) can be used, since all sides are known. You can form opp/hyp for sin⊃⁻¹, adj/hyp for cos⊃⁻¹, or opp/adj for tan⊃⁻¹. All give the same angle — you have three options.

9. Misconception correction

   1. Student A: cos(22°) = 9/23: Error: the sides 9 (opp) and 23 (hyp) belong to sine, not cosine. Correct: sin(22°) = 9/23.
   2. Student B: sin(67°) = 12/5: Error: sin uses opp/hyp, but 12 (opp) and 5 (adj) are not opp and hyp — they are opp and adj. Correct: tan(67°) = 12/5.
   3. Student C: tan(40°) = 7/11: Error: the sides 7 (adj) and 11 (hyp) belong to cosine. Correct: cos(40°) = 7/11.

10. Real context: trigonometry from a map

    1. sin θ, cos θ, tan θ (hyp unknown): tan θ = opp/adj = 30/50 = 3/5. sin θ = 30/hyp and cos θ = 50/hyp (hyp not yet known).
    2. Exact hypotenuse: hyp = √(30²+50²) = √(900+2500) = √3400 = 10√34 ≈ 58.31 km
    3. All three ratios with hyp: sin θ = 30/(10√34) = 3/√34 ≈ 0.514. cos θ = 50/(10√34) = 5/√34 ≈ 0.857. tan θ = 3/5 = 0.6 (unchanged).