Practice Maths

Solutions: The Cosine Rule

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  1. Fluency

    1. a = 7, b = 10, C = 45°.

      \[c^2 = 7^2 + 10^2 - 2(7)(10)\cos 45° = 49 + 100 - 140 \times 0.7071 = 149 - 98.99 = 50.01\] \[c = \sqrt{50.01} \approx \mathbf{7.07}\]

    2. a = 15, b = 20, C = 110°.

      \[c^2 = 225 + 400 - 2(15)(20)\cos 110° = 625 - 600 \times (-0.3420) = 625 + 205.2 = 830.2\] \[c = \sqrt{830.2} \approx \mathbf{28.81}\]

    3. b = 8, c = 12, A = 72°. Here A is the included angle between b and c.

      \[a^2 = b^2 + c^2 - 2bc\cos A = 64 + 144 - 2(8)(12)\cos 72°\] \[= 208 - 192 \times 0.3090 = 208 - 59.33 = 148.67\] \[a = \sqrt{148.67} \approx \mathbf{12.19}\]
  2. Fluency

    1. a = 5, b = 8, c = 10. Find C (largest angle, opposite c = 10).

      \[\cos C = \frac{25 + 64 - 100}{2(5)(8)} = \frac{-11}{80} = -0.1375\] \[C = \cos^{-1}(-0.1375) \approx \mathbf{97.9°}\]

    2. a = 12, b = 9, c = 15. Largest angle C (opposite c = 15).

      \[\cos C = \frac{144 + 81 - 225}{2(12)(9)} = \frac{0}{216} = 0\] \[C = 90°\]

      (This is a right-angled triangle! 9–12–15 is a 3–4–5 scaled by 3.)

      \[\cos A = \frac{81 + 225 - 144}{2(9)(15)} = \frac{162}{270} = 0.6\implies A = \cos^{-1}(0.6) \approx 53.1°\] \[B = 180° - 90° - 53.1° = 36.9°\]

      Angles: A ≈ 53.1°, B ≈ 36.9°, C = 90°

    3. a = 6, b = 6, c = 9. The angle between the equal sides (a and b) is C.

      \[\cos C = \frac{36 + 36 - 81}{2(6)(6)} = \frac{-9}{72} = -0.125\] \[C = \cos^{-1}(-0.125) \approx \mathbf{97.2°}\]
  3. Fluency

    Sides 14 cm, 18 cm, 22 cm. Let a = 14, b = 18, c = 22. Find largest angle C first.

    \[\cos C = \frac{196 + 324 - 484}{2(14)(18)} = \frac{36}{504} = 0.07143\] \[C = \cos^{-1}(0.07143) \approx 85.9°\] \[\cos A = \frac{324 + 484 - 196}{2(18)(22)} = \frac{612}{792} = 0.7727\] \[A \approx \cos^{-1}(0.7727) \approx 39.5°\] \[B = 180° - 85.9° - 39.5° = 54.6°\]

    Angles: A ≈ 39.5°, B ≈ 54.6°, C ≈ 85.9°

  4. Understanding

    Forces 50 N and 70 N at 58°. In the force triangle, the angle between the two sides (50 N and 70 N) is the supplement: 180° − 58° = 122°.

    \[R^2 = 50^2 + 70^2 - 2(50)(70)\cos 122°\] \[= 2500 + 4900 - 7000 \times (-0.5299)\] \[= 7400 + 3709.3 = 11109.3\] \[R = \sqrt{11109.3} \approx \mathbf{105.4 \text{ N}}\]
  5. Understanding

    Leg 1: 45 km on bearing 030°. Leg 2: 60 km on bearing 135°.

    The change in direction: 135° − 030° = 105°. This is the exterior angle at the turning point. The interior angle between the two legs = 180° − 105° = 75°.

    \[d^2 = 45^2 + 60^2 - 2(45)(60)\cos 75°\] \[= 2025 + 3600 - 5400 \times 0.2588\] \[= 5625 - 1397.5 = 4227.5\] \[d = \sqrt{4227.5} \approx \mathbf{65.02 \text{ km}}\]
  6. Understanding

    1. Let a = 85, b = 110, c = 140. Find largest angle C first.

      \[\cos C = \frac{85^2 + 110^2 - 140^2}{2(85)(110)} = \frac{7225 + 12100 - 19600}{18700} = \frac{-275}{18700} = -0.01471\] \[C = \cos^{-1}(-0.01471) \approx 90.8°\] \[\cos A = \frac{110^2 + 140^2 - 85^2}{2(110)(140)} = \frac{12100 + 19600 - 7225}{30800} = \frac{24475}{30800} = 0.7947\] \[A \approx 37.4°\] \[B = 180° - 90.8° - 37.4° = 51.8°\]

      A ≈ 37.4°, B ≈ 51.8°, C ≈ 90.8°

    2. Largest angle: C ≈ 90.8° (opposite the 140 m side).

    3. Area = ½ab sin C = ½ × 85 × 110 × sin 90.8°

      \[= \tfrac{1}{2} \times 85 \times 110 \times 0.9999 \approx \mathbf{4674.5 \text{ m}^2}\]

      Since C ≈ 90°, the area ≈ ½ × 85 × 110 = 4675 m² as a check. ✓

  7. Understanding

    1. Method: Sine rule (SSA — two sides and angle opposite one of them).

      B = 55°, a = 14, b = 18. Find A.

      \[\sin A = \frac{a \sin B}{b} = \frac{14 \times \sin 55°}{18} = \frac{14 \times 0.8192}{18} = \frac{11.469}{18} \approx 0.6372\] \[A \approx \sin^{-1}(0.6372) \approx \mathbf{39.5°}\]

      Check ambiguous: A2 = 140.5°. B + A2 = 55 + 140.5 = 195.5° > 180°. Only one solution.

    2. Method: Cosine rule (SAS — two sides and included angle C = 47°).

      \[c^2 = 64 + 144 - 2(8)(12)\cos 47° = 208 - 192 \times 0.6820 = 208 - 130.94 = 77.06\] \[c = \sqrt{77.06} \approx \mathbf{8.78}\]

    3. Method: Sine rule (AAS — two angles and a side).

      A = 32°, B = 69°, C = 180 − 32 − 69 = 79°. Side c = 15 is opposite C = 79°.

      \[a = \frac{c \sin A}{\sin C} = \frac{15 \times \sin 32°}{\sin 79°} = \frac{15 \times 0.5299}{0.9816} \approx \mathbf{8.10}\]
  8. Understanding

    1. Triangle ABF (F = fire): AB = 12 km, angle at A = 52°, angle at B = 74°, angle at F = 180 − 52 − 74 = 54°.

      \[AF = \frac{12 \times \sin 74°}{\sin 54°} = \frac{12 \times 0.9613}{0.8090} \approx \mathbf{14.25 \text{ km}}\] \[BF = \frac{12 \times \sin 52°}{\sin 54°} = \frac{12 \times 0.7880}{0.8090} \approx \mathbf{11.68 \text{ km}}\]

    2. Using cosine rule to find AB from AF and BF: if the answer above is correct, then:

      \[AB^2 = AF^2 + BF^2 - 2 \cdot AF \cdot BF \cdot \cos(\angle AFB)\] \[= 14.25^2 + 11.68^2 - 2(14.25)(11.68)\cos 54°\] \[= 203.06 + 136.42 - 332.95 \times 0.5878\] \[= 339.48 - 195.8 = 143.68\] \[AB = \sqrt{143.68} \approx 11.99 \text{ km} \approx 12 \text{ km} \checkmark\]
  9. Problem Solving

    1. a = 280, b = 350, C = 120°.

      \[c^2 = 280^2 + 350^2 - 2(280)(350)\cos 120°\] \[= 78400 + 122500 - 196000 \times (-0.5)\] \[= 200900 + 98000 = 298900\] \[c = \sqrt{298900} \approx \mathbf{546.7 \text{ m}}\]

    2. Find angle A (opposite a = 280):

      \[\cos A = \frac{350^2 + 546.7^2 - 280^2}{2(350)(546.7)} = \frac{122500 + 298881 - 78400}{382690} = \frac{342981}{382690} \approx 0.8963\] \[A \approx 26.4°\] \[B = 180° - 120° - 26.4° = 33.6°\]

      Angles: A ≈ 26.4°, B ≈ 33.6°, C = 120°

    3. \[\text{Area} = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(280)(350)\sin 120° = 49000 \times 0.8660 \approx \mathbf{42{,}434 \text{ m}^2}\]

    4. Perimeter = 280 + 350 + 546.7 = 1176.7 m

  10. Problem Solving

    1. PQ = 45, QR = 72, PR = 61. Find angle QPR (= angle P).

      \[\cos P = \frac{PQ^2 + PR^2 - QR^2}{2 \cdot PQ \cdot PR} = \frac{45^2 + 61^2 - 72^2}{2(45)(61)}\] \[= \frac{2025 + 3721 - 5184}{5490} = \frac{562}{5490} \approx 0.10237\] \[P = \cos^{-1}(0.10237) \approx \mathbf{84.1°}\]

    2. The perpendicular from Q to PR has length h, where in right triangle QPD (D = foot of perpendicular on PR):

      \[h = PQ \sin P = 45 \times \sin 84.1° = 45 \times 0.9947 \approx \mathbf{44.76 \text{ km}}\]

    3. Distance from P to foot of perpendicular D:

      \[PD = PQ \cos P = 45 \times \cos 84.1° = 45 \times 0.1024 \approx \mathbf{4.61 \text{ km from P}}\]

      The perpendicular road meets PR approximately 4.61 km from town P.