Practice Maths

Solutions: The Sine Rule

← Back to Questions›Applications of Trigonometry

  1. Fluency

    1. A = 55°, B = 75°, a = 12. Find b.

      \[b = \frac{a \sin B}{\sin A} = \frac{12 \times \sin 75°}{\sin 55°} = \frac{12 \times 0.9659}{0.8192} \approx \mathbf{14.14}\]

    2. B = 40°, C = 65°, b = 18. Find c.

      \[c = \frac{b \sin C}{\sin B} = \frac{18 \times \sin 65°}{\sin 40°} = \frac{18 \times 0.9063}{0.6428} \approx \mathbf{25.37}\]

    3. A = 36°, C = 82°, c = 25. Find a.

      \[a = \frac{c \sin A}{\sin C} = \frac{25 \times \sin 36°}{\sin 82°} = \frac{25 \times 0.5878}{0.9903} \approx \mathbf{14.84}\]
  2. Fluency

    1. a = 9, b = 13, A = 38°. Find B.

      \[\sin B = \frac{b \sin A}{a} = \frac{13 \times \sin 38°}{9} = \frac{13 \times 0.6157}{9} = \frac{8.004}{9} \approx 0.8894\] \[B = \sin^{-1}(0.8894) \approx \mathbf{62.9°}\]

      Check ambiguous: B2 = 180° − 62.9° = 117.1°. Check: A + B2 = 38 + 117.1 = 155.1° < 180°. So B2 = 117.1° is also valid. Two triangles exist. Both B ≈ 62.9° and B ≈ 117.1° are valid.

    2. a = 18, c = 22, C = 65°. Find A.

      \[\sin A = \frac{a \sin C}{c} = \frac{18 \times \sin 65°}{22} = \frac{18 \times 0.9063}{22} = \frac{16.31}{22} \approx 0.7414\] \[A = \sin^{-1}(0.7414) \approx 47.8°\]

      Check A2: 180 − 47.8 = 132.2°. C + A2 = 65 + 132.2 = 197.2° > 180°. Invalid. One solution: A ≈ 47.8°

    3. b = 14, c = 11, B = 82°. Find C.

      \[\sin C = \frac{c \sin B}{b} = \frac{11 \times \sin 82°}{14} = \frac{11 \times 0.9903}{14} = \frac{10.89}{14} \approx 0.7782\] \[C = \sin^{-1}(0.7782) \approx 51.1°\]

      Check C2: 180 − 51.1 = 128.9°. B + C2 = 82 + 128.9 = 210.9° > 180°. Invalid. One solution: C ≈ 51.1°

  3. Fluency

    A = 48°, B = 67°, b = 21.

    Third angle: C = 180° − 48° − 67° = 65°

    Find a:

    \[a = \frac{b \sin A}{\sin B} = \frac{21 \times \sin 48°}{\sin 67°} = \frac{21 \times 0.7431}{0.9205} \approx \mathbf{16.94}\]

    Find c:

    \[c = \frac{b \sin C}{\sin B} = \frac{21 \times \sin 65°}{\sin 67°} = \frac{21 \times 0.9063}{0.9205} \approx \mathbf{20.67}\]
  4. Understanding

    a = 10, b = 14, A = 44°. Find B.

    \[\sin B = \frac{b \sin A}{a} = \frac{14 \times \sin 44°}{10} = \frac{14 \times 0.6947}{10} = \frac{9.726}{10} = 0.9726\] \[B_1 = \sin^{-1}(0.9726) \approx 76.6°\] \[B_2 = 180° - 76.6° = 103.4°\]

    Check B1: A + B1 = 44 + 76.6 = 120.6° < 180° ✓  C1 = 59.4°

    Check B2: A + B2 = 44 + 103.4 = 147.4° < 180° ✓  C2 = 32.6°

    Two triangles are possible:

    Triangle 1: B ≈ 76.6°, C ≈ 59.4°

    Triangle 2: B ≈ 103.4°, C ≈ 32.6°

  5. Understanding

    1. Angles: C = 74°, A = 56°.   B = 180° − 74° − 56° = 50°

    2. CA is side b (opposite B = 50°). AB = c = 85 m (opposite C = 74°).

      \[b = CA = \frac{c \sin B}{\sin C} = \frac{85 \times \sin 50°}{\sin 74°} = \frac{85 \times 0.7660}{0.9613} \approx \mathbf{67.7 \text{ m}}\]

    3. CB is side a (opposite A = 56°).

      \[a = CB = \frac{c \sin A}{\sin C} = \frac{85 \times \sin 56°}{\sin 74°} = \frac{85 \times 0.8290}{0.9613} \approx \mathbf{73.3 \text{ m}}\]
  6. Understanding

    In triangle formed by port (P), Ship 1 (S1), Ship 2 (S2):

    PS1 = 80 km, S1S2 = 95 km, angle at P = 70°.

    Use sine rule: S1S2/sin(angle at P) = PS1/sin(angle at S2)

    \[\frac{95}{\sin 70°} = \frac{80}{\sin(\angle S_2)}\] \[\sin(\angle S_2) = \frac{80 \times \sin 70°}{95} = \frac{80 \times 0.9397}{95} = \frac{75.17}{95} \approx 0.7913\] \[\angle S_2 = \sin^{-1}(0.7913) \approx 52.2°\]

    Angle at S1 = 180° − 70° − 52.2° = 57.8°

    \[PS_2 = \frac{S_1S_2 \times \sin(\angle S_1)}{\sin(\angle P)} = \frac{95 \times \sin 57.8°}{\sin 70°} = \frac{95 \times 0.8450}{0.9397} \approx \mathbf{85.4 \text{ km}}\]
  7. Understanding

    1. Base angles 52° and 63°. Apex = 180° − 52° − 63° = 65°

    2. Base c = 14 m (opposite apex C = 65°).

      Side a (opposite A = 52°):

      \[a = \frac{14 \times \sin 52°}{\sin 65°} = \frac{14 \times 0.7880}{0.9063} \approx \mathbf{12.17 \text{ m}}\]

      Side b (opposite B = 63°):

      \[b = \frac{14 \times \sin 63°}{\sin 65°} = \frac{14 \times 0.8910}{0.9063} \approx \mathbf{13.76 \text{ m}}\]

    3. Side b = 13.76 m > a = 12.17 m. Side b is longer by 13.76 − 12.17 = 1.59 m.

      Note: the longer sloping side is opposite the larger base angle (63° > 52°), as expected.

  8. Understanding

    B is due east of A; AB = 600 m. Bearing of P from A = N 25° E, so angle PAB = 25° (measured from North, which is perpendicular to AB). Since AB runs east-west, the angle at A in triangle ABP is 90° − 25° = 65° from AB.

    Bearing of P from B = N 60° W. Since AB runs west-to-east, N 60° W from B makes an angle of 90° + 60° = 150° from BA, so the interior angle at B = 180° − 150° = 30°...

    More carefully: angle ABP at B = 90° − 60° = 30° below north, which is 90° + 60° from the east direction. Interior angle at B (measured inside triangle ABP) = 180° − (90° + 60°) ...

    Let’s use the compass angles directly. At A: P is N 25° E, AB points East (090°). Angle between AB and AP = 90° − 25° = 65° (angle PAB inside triangle).

    At B: P is N 60° W (bearing 300°), BA points West (270°). Angle between BA and BP = 300° − 270° = 30° (angle ABP inside triangle).

    Angle APB = 180° − 65° − 30° = 85°

    \[AP = \frac{AB \sin(\angle ABP)}{\sin(\angle APB)} = \frac{600 \times \sin 30°}{\sin 85°} = \frac{600 \times 0.5}{0.9962} \approx \mathbf{301.1 \text{ m}}\] \[BP = \frac{AB \sin(\angle PAB)}{\sin(\angle APB)} = \frac{600 \times \sin 65°}{\sin 85°} = \frac{600 \times 0.9063}{0.9962} \approx \mathbf{545.9 \text{ m}}\]
  9. Problem Solving

    1. Known side c = 320 m (base), angles at each end: C’s opposite-to-base angles are the two base angles. Let A = 61°, B = 74°, c = 320 m.

      Third angle: C = 180° − 61° − 74° = 45°

      \[a = \frac{c \sin A}{\sin C} = \frac{320 \times \sin 61°}{\sin 45°} = \frac{320 \times 0.8746}{0.7071} \approx \mathbf{395.4 \text{ m}}\] \[b = \frac{c \sin B}{\sin C} = \frac{320 \times \sin 74°}{\sin 45°} = \frac{320 \times 0.9613}{0.7071} \approx \mathbf{434.8 \text{ m}}\]

    2. Perimeter = 320 + 395.4 + 434.8 = 1150.2 m

    3. Height from C to base c = 320 m: using side a and angle B, the perpendicular height h from apex to base:

      \[h = b \sin A = 434.8 \times \sin 61° = 434.8 \times 0.8746 \approx 380.3 \text{ m}\]

      Verify: h = a sin B = 395.4 × sin 74° = 395.4 × 0.9613 ≈ 380.1 m ✓

      \[\text{Area} = \frac{1}{2} \times 320 \times 380.3 \approx \mathbf{60\,848 \text{ m}^2}\]

      (Or use Area = ½ab sin C = ½ × 395.4 × 434.8 × sin 45° ≈ 60,848 m² ✓)

  10. Problem Solving AB = 8 cm (= c), BC = 6 cm (= a), angle BAC = A = 35°.

    1. Find angle BCA = C. Use sine rule: sin C / c = sin A / a → but here we need: sin C / AB = sin A / BC.

      Side AB = c = 8 is opposite C, side BC = a = 6 is opposite A = 35°.

      \[\sin C = \frac{c \sin A}{a} = \frac{8 \times \sin 35°}{6} = \frac{8 \times 0.5736}{6} = \frac{4.589}{6} \approx 0.7648\] \[C_1 = \sin^{-1}(0.7648) \approx 49.9°\]

    2. C2 = 180° − 49.9° = 130.1°.

      Check: A + C2 = 35 + 130.1 = 165.1° < 180° ✓

      The ambiguous case applies: two triangles are possible.

      Triangle 1: C1 ≈ 49.9°, B1 = 180 − 35 − 49.9 = 95.1°

      Triangle 2: C2 ≈ 130.1°, B2 = 180 − 35 − 130.1 = 14.9°

    3. Find AC = b (opposite B) using sine rule b/sin B = a/sin A:

      Triangle 1: B1 = 95.1°

      \[b_1 = \frac{a \sin B_1}{\sin A} = \frac{6 \times \sin 95.1°}{\sin 35°} = \frac{6 \times 0.9961}{0.5736} \approx \mathbf{10.41 \text{ cm}}\]

      Triangle 2: B2 = 14.9°

      \[b_2 = \frac{6 \times \sin 14.9°}{\sin 35°} = \frac{6 \times 0.2571}{0.5736} \approx \mathbf{2.69 \text{ cm}}\]

    4. Area = ½ab sin C (using the two known sides a = BC = 6, b = AC, and included angle B).

      Alternatively, use sides a and c with included angle B:

      Triangle 1:

      \[\text{Area}_1 = \tfrac{1}{2} \times a \times c \times \sin B_1 = \tfrac{1}{2} \times 6 \times 8 \times \sin 95.1° = 24 \times 0.9961 \approx \mathbf{23.91 \text{ cm}^2}\]

      Triangle 2:

      \[\text{Area}_2 = \tfrac{1}{2} \times 6 \times 8 \times \sin 14.9° = 24 \times 0.2571 \approx \mathbf{6.17 \text{ cm}^2}\]