Solutions: Right-Angle Triangle Trigonometry

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1. Fluency Find the marked side, correct to 2 decimal places.
   1. Show Answer

      θ = 42°, H = 18 cm, want O. Use sin.

      \[\sin 42° = \frac{O}{18}\] \[O = 18 \times \sin 42° = 18 \times 0.6691\ldots \approx \mathbf{12.04 \text{ cm}}\]
   2. Show Answer

      θ = 67°, A = 11 m, want H. Use cos.

      \[\cos 67° = \frac{11}{H}\] \[H = \frac{11}{\cos 67°} = \frac{11}{0.3907\ldots} \approx \mathbf{28.15 \text{ m}}\]
   3. Show Answer

      θ = 28°, H = 25 cm, want A. Use cos.

      \[\cos 28° = \frac{A}{25}\] \[A = 25 \times \cos 28° = 25 \times 0.8829\ldots \approx \mathbf{22.07 \text{ cm}}\]
2. Fluency Find the marked angle to the nearest degree.
   1. Show Answer

      O = 7, H = 15. Use sin⁻¹.

      \[\theta = \sin^{-1}\!\left(\frac{7}{15}\right) = \sin^{-1}(0.4\overline{6}) \approx \mathbf{28°}\]
   2. Show Answer

      A = 8, H = 13. Use cos⁻¹.

      \[\theta = \cos^{-1}\!\left(\frac{8}{13}\right) = \cos^{-1}(0.6154\ldots) \approx \mathbf{52°}\]
   3. Show Answer

      O = 12, A = 9. Use tan⁻¹.

      \[\theta = \tan^{-1}\!\left(\frac{12}{9}\right) = \tan^{-1}(1.\overline{3}) \approx \mathbf{53°}\]
3. Fluency Right triangle with legs 6 cm and 10 cm.
   1. Show Answer

      By Pythagoras:

      \[H = \sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136} \approx \mathbf{11.66 \text{ cm}}\]
   2. Show Answer

      Let α be the angle opposite the 6 cm leg:

      \[\alpha = \tan^{-1}\!\left(\frac{6}{10}\right) = \tan^{-1}(0.6) \approx 31°\]

      The other angle: β = 90° − 31° = 59°

      The two acute angles are approximately 31° and 59°.

4. Understanding Ladder 8 m, base 2.5 m from wall.
   1. Show Answer

      The ladder is the hypotenuse (H = 8 m). The base distance is adjacent (A = 2.5 m). Want angle θ with ground.

      \[\cos\theta = \frac{A}{H} = \frac{2.5}{8} = 0.3125\] \[\theta = \cos^{-1}(0.3125) \approx \mathbf{71.8°}\]
   2. Show Answer

      Height up the wall = opposite side O.

      \[O = \sqrt{H^2 - A^2} = \sqrt{64 - 6.25} = \sqrt{57.75} \approx \mathbf{7.60 \text{ m}}\]

      Or: O = H sin θ = 8 × sin 71.8° ≈ 7.60 m. ✓

5. Understanding Ramp rises 1.2 m over 6 m horizontal.
   1. Show Answer

      Rise = O = 1.2 m, horizontal = A = 6 m. Use tan⁻¹.

      \[\theta = \tan^{-1}\!\left(\frac{1.2}{6}\right) = \tan^{-1}(0.2) \approx \mathbf{11.3°}\]
   2. Show Answer

      Length of ramp surface = hypotenuse H.

      \[H = \sqrt{1.2^2 + 6^2} = \sqrt{1.44 + 36} = \sqrt{37.44} \approx \mathbf{6.12 \text{ m}}\]
6. Understanding Show Answer

   Horizontal distance A = 50 m, angle of elevation θ = 38°. Want height O.

   \[\tan 38° = \frac{O}{50}\] \[O = 50 \times \tan 38° = 50 \times 0.7813\ldots \approx \mathbf{39.06 \text{ m}}\]
7. Understanding Pole 12 m high, base 9 m from anchor.
   1. Show Answer

      O = 12 m (pole height), A = 9 m (base distance). Length of wire = H.

      \[H = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = \mathbf{15 \text{ m}}\]
   2. Show Answer

      Angle θ with ground: tanθ = O/A = 12/9

      \[\theta = \tan^{-1}\!\left(\frac{12}{9}\right) = \tan^{-1}(1.\overline{3}) \approx \mathbf{53.1°}\]
8. Understanding Rectangular park 80 m × 60 m, diagonal path.
   1. Show Answer

      The diagonal forms a right triangle with legs 80 m and 60 m.

      \[\text{Length} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = \mathbf{100 \text{ m}}\]
   2. Show Answer

      The opposite side to θ (measured from the 80 m side) is 60 m. Adjacent = 80 m.

      \[\theta = \tan^{-1}\!\left(\frac{60}{80}\right) = \tan^{-1}(0.75) \approx \mathbf{36.9°}\]
9. Problem Solving Triangle ABC, right angle at C, AC = 7 cm, BC = 24 cm.
   1. Show Answer

      AB is the hypotenuse (opposite right angle at C).

      \[AB = \sqrt{AC^2 + BC^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = \mathbf{25 \text{ cm}}\]
   2. Show Answer

      Angle ABC: the side opposite to B is AC = 7, hypotenuse AB = 25.

      \[\sin(\angle ABC) = \frac{AC}{AB} = \frac{7}{25} = 0.28\] \[\angle ABC = \sin^{-1}(0.28) \approx \mathbf{16°}\]
   3. Show Answer

      Angle BAC: side opposite A is BC = 24, hypotenuse AB = 25.

      \[\sin(\angle BAC) = \frac{BC}{AB} = \frac{24}{25} = 0.96\] \[\angle BAC = \sin^{-1}(0.96) \approx \mathbf{74°}\]
   4. Show Answer

      Angle sum: 90° + 16° + 74° = 180° ✓

      (Exact values: sin&sup8;&sup9;ABC = 7/25, so ∠ABC = 16.26°; ∠BAC = 73.74°. Sum = 90 + 16.26 + 73.74 = 180.00°.)

10. Problem Solving Two poles 20 m apart: Pole A = 8 m, Pole B = 15 m. Wires cross at X.
    1. Show Answer

       Wire from A-top to B-base: rises 8 m over 20 m. Angle with ground = tan⁻¹(8/20) ≈ 21.8°

       Wire from B-top to A-base: rises 15 m over 20 m. Angle with ground = tan⁻¹(15/20) ≈ 36.9°

    2. Show Answer

       Let d = horizontal distance from Pole A to crossing point X.

       Wire from A-top to B-base descends from 8 m at Pole A to 0 m at Pole B (20 m away). Height at position d from A:

       \[h_1 = 8 - \frac{8d}{20} = 8 - 0.4d\]

       Wire from B-top to A-base descends from 15 m at Pole B to 0 m at Pole A. Height at position d from A:

       \[h_2 = \frac{15d}{20} = 0.75d\]

       At crossing point X: h₁ = h₂

       \[8 - 0.4d = 0.75d\] \[8 = 1.15d\] \[d = \frac{8}{1.15} \approx 6.96 \text{ m from Pole A}\]
    3. Show Answer

       Height of crossing point:

       \[h = 0.75d = 0.75 \times \frac{8}{1.15} = \frac{6}{1.15} \approx \mathbf{5.22 \text{ m}}\]

       Verify: h₁ = 8 − 0.4(6.96) = 8 − 2.78 = 5.22 m ✓

       The crossing point is approximately 5.22 m above the ground.

       Note: This result (h = ab/(a+b) where a and b are the pole heights) generalises to any two poles — a beautiful result from similar triangles!