Practice Maths

Unit 2 Topic 2 Review — Applications of Trigonometry — Solutions

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This review covers all five lessons in Applications of Trigonometry: Right-Angle Trigonometry, Angles of Elevation, Depression and Bearings, The Sine Rule, The Cosine Rule, and Area of a Triangle. Allow approximately 60–75 minutes for this review.

Review Questions

  1. Q1 — Right-angle trigonometry

    Fluency

    In a right-angled triangle, one angle is 38° and the hypotenuse is 22 cm.

    (a) Find the side opposite the 38° angle.

    (b) Find the side adjacent to the 38° angle.

    (a) Opposite side:

    sin(38°) = opp ÷ hyp

    opp = 22 × sin(38°) = 22 × 0.6157 = 13.55 cm

    (b) Adjacent side:

    cos(38°) = adj ÷ hyp

    adj = 22 × cos(38°) = 22 × 0.7880 = 17.34 cm

    Check: opp² + adj² = 13.55² + 17.34² = 183.6 + 300.7 = 484.3 ≈ 22² = 484 ✓

  2. Q2 — Angle of elevation

    Fluency

    From a point 65 m from the base of a mast (on level ground), the top of the mast is observed at an angle of elevation of 42°. Find the height of the mast.

    Using right-angle trigonometry:

    tan(42°) = height ÷ 65

    height = 65 × tan(42°)

    height = 65 × 0.9004

    height = 58.53 m

  3. Q3 — Sine rule to find sides

    Fluency

    In triangle ABC: angle A = 58°, angle B = 75°, and side a = 18 cm.

    (a) Find side b.

    (b) Find side c (note: C = 180° − 58° − 75° = 47°).

    Angle C = 180 − 58 − 75 = 47°

    (a) Finding side b using the sine rule:

    b/sin B = a/sin A

    b/sin(75°) = 18/sin(58°)

    b = 18 × sin(75°) ÷ sin(58°)

    b = 18 × 0.9659 ÷ 0.8480

    b = 20.50 cm

    (b) Finding side c:

    c/sin C = a/sin A

    c/sin(47°) = 18/sin(58°)

    c = 18 × sin(47°) ÷ sin(58°)

    c = 18 × 0.7314 ÷ 0.8480

    c = 15.52 cm

  4. Q4 — Cosine rule to find a side

    Fluency

    In triangle ABC: a = 11 m, b = 16 m and angle C = 64°. Find side c.

    Using the cosine rule: c² = a² + b² − 2ab cos C

    c² = 11² + 16² − 2 × 11 × 16 × cos(64°)

    c² = 121 + 256 − 352 × 0.4384

    c² = 377 − 154.32

    c² = 222.68

    c = √222.68 = 14.92 m

  5. Q5 — Area of a triangle

    Fluency

    A triangular garden has two sides of 14 m and 20 m with an included angle of 55°. Find the area of the garden.

    Using: Area = ½ ab sin C

    Area = ½ × 14 × 20 × sin(55°)

    Area = 140 × 0.8192

    Area = 114.69 m²

  6. Q6 — Ladder against a wall

    Understanding

    A ladder 5.5 m long leans against a vertical wall, making an angle of 68° with the horizontal ground.

    (a) Find the height the ladder reaches on the wall.

    (b) Find the distance of the base of the ladder from the wall.

    The ladder is the hypotenuse (5.5 m); the angle with the ground is 68°.

    (a) Height reached (opposite side):

    sin(68°) = height ÷ 5.5

    height = 5.5 × sin(68°) = 5.5 × 0.9272 = 5.10 m

    (b) Distance from wall (adjacent side):

    cos(68°) = distance ÷ 5.5

    distance = 5.5 × cos(68°) = 5.5 × 0.3746 = 2.06 m

  7. Q7 — Angles of depression to two boats

    Understanding

    From the top of a 120 m cliff, two boats are seen on the same side. The first boat is at an angle of depression of 28° and the second at 42°.

    (a) Find the horizontal distance to the first boat.

    (b) Find the horizontal distance to the second boat.

    (c) Find the distance between the two boats.

    The cliff is vertical (120 m). Angle of depression = angle of elevation from boat to cliff top.

    (a) Distance to first boat (28° depression):

    tan(28°) = 120 ÷ d1

    d1 = 120 ÷ tan(28°) = 120 ÷ 0.5317 = 225.7 m

    (b) Distance to second boat (42° depression):

    tan(42°) = 120 ÷ d2

    d2 = 120 ÷ tan(42°) = 120 ÷ 0.9004 = 133.3 m

    (c) Distance between the two boats:

    Both boats are on the same side, so:

    Distance = d1 − d2 = 225.7 − 133.3 = 92.4 m

  8. Q8 — Bearing and displacement

    Understanding

    A ship sails from port on a bearing of 125° for 75 km.

    (a) How far south of port is the ship?

    (b) How far east of port is the ship?

    Bearing 125° means 125° measured clockwise from North.

    This is equivalent to S 55° E (or 35° from the south axis).

    The angle from South = 125° − 90° = 35° (angle measured from East)

    Alternatively: angle from North axis towards East = 125°.

    South component = 75 × cos(125° − 90°) ... using standard decomposition:

    Southward (y-component): 75 × cos(180° − 125°) ...

    Direct approach: The bearing 125° makes an angle of (125° − 90°) = 35° south of east.

    (a) Southward distance:

    South = 75 × sin(125° − 90°) ...

    Using bearing components: North component = 75 cos(125°) = 75 × (−0.5736) = −43.02 km

    (Negative means southward) → 43.0 km south

    (b) Eastward distance:

    East component = 75 sin(125°) = 75 × 0.8192 = 61.4 km east

  9. Q9 — Sine rule in surveying

    Understanding

    A surveyor stands at point C and observes two points A and B. The angle ACB = 68°, angle CAB = 54°, and the distance AB = 120 m.

    (a) Find the distance CA.

    (b) Find the distance CB.

    First find angle CBA = 180° − 68° − 54° = 58°

    The side AB is opposite angle C (= 68°), so AB = c = 120 m, angle C = 68°.

    Angle A = 54° (opposite side CB = a), angle B = 58° (opposite side CA = b).

    (a) Finding CA (= b, opposite angle B = 58°):

    b/sin B = c/sin C

    CA/sin(58°) = 120/sin(68°)

    CA = 120 × sin(58°) ÷ sin(68°)

    CA = 120 × 0.8480 ÷ 0.9272

    CA = 109.76 m

    (b) Finding CB (= a, opposite angle A = 54°):

    CB/sin(54°) = 120/sin(68°)

    CB = 120 × sin(54°) ÷ sin(68°)

    CB = 120 × 0.8090 ÷ 0.9272

    CB = 104.7 m

  10. Q10 — Cosine rule to find the smallest angle

    Understanding

    A triangle has sides of 9 m, 15 m and 20 m. Find the smallest angle of the triangle.

    The smallest angle is opposite the shortest side (9 m).

    Let a = 9, b = 15, c = 20. Find angle A (opposite a = 9).

    Cosine rule: cos A = (b² + c² − a²) ÷ (2bc)

    cos A = (15² + 20² − 9²) ÷ (2 × 15 × 20)

    cos A = (225 + 400 − 81) ÷ 600

    cos A = 544 ÷ 600 = 0.9067

    A = cos−1(0.9067) = 25.0°

    The smallest angle is 25.0°.

  11. Q11 — Area then perimeter

    Understanding

    A triangle has two sides of 18 cm and 25 cm with an included angle of 72°.

    (a) Find the area of the triangle.

    (b) Find the length of the third side.

    (c) Find the perimeter.

    (a) Area:

    Area = ½ × 18 × 25 × sin(72°)

    Area = 225 × sin(72°) = 225 × 0.9511

    Area = 213.99 cm²

    (b) Third side (cosine rule):

    Let a = 18, b = 25, C = 72°

    c² = 18² + 25² − 2 × 18 × 25 × cos(72°)

    c² = 324 + 625 − 900 × 0.3090

    c² = 949 − 278.1 = 670.9

    c = √670.9 = 25.90 cm

    (c) Perimeter:

    Perimeter = 18 + 25 + 25.90 = 68.90 cm

  12. Q12 — Triangular paddock

    Problem Solving

    A triangular paddock has two sides of 280 m and 350 m with an included angle of 62°.

    (a) Find the length of the third side.

    (b) Find all three angles of the paddock.

    (c) Find the area of the paddock.

    (d) Find the cost to fence the entire paddock at $15 per metre.

    (a) Third side (cosine rule):

    Let a = 280, b = 350, C = 62°

    c² = 280² + 350² − 2 × 280 × 350 × cos(62°)

    c² = 78 400 + 122 500 − 196 000 × 0.4695

    c² = 200 900 − 92 022 = 108 878

    c = √108 878 = 329.97 m ≈ 330.0 m

    (b) Finding remaining angles:

    Use sine rule to find angle A (opposite a = 280):

    sin A / 280 = sin(62°) / 330.0

    sin A = 280 × sin(62°) ÷ 330.0 = 280 × 0.8829 ÷ 330.0 = 0.7487

    A = sin−1(0.7487) = 48.5°

    Angle B = 180 − 62 − 48.5 = 69.5°

    (c) Area:

    Area = ½ × 280 × 350 × sin(62°)

    = 49 000 × 0.8829 = 43 262 m²

    (d) Fencing cost:

    Perimeter = 280 + 350 + 330.0 = 960.0 m

    Cost = 960.0 × $15 = $14 400

  13. Q13 — Navigation problem with two legs

    Problem Solving

    A ship sails 40 km on a bearing of 050°, then turns and sails 65 km on a bearing of 140°.

    (a) Find the angle between the two legs of the journey.

    (b) Find the direct distance from the starting point to the final position.

    (c) Find the bearing to sail directly back to the starting port.

    (a) Angle between the two legs:

    The ship sails on bearing 050° then turns to bearing 140°.

    The exterior angle at the turning point = 140° − 050° = 90°.

    The interior angle between the two legs (inside the triangle) = 180° − 90° = 90°.

    Verification: The supplementary angle at the vertex (between the direction of travel on leg 1 and leg 2) = 140° − 50° = 90°. So the two legs are perpendicular.

    (b) Direct distance (Pythagoras, since angle = 90°):

    d² = 40² + 65² = 1600 + 4225 = 5825

    d = √5825 = 76.32 km

    (c) Bearing to return to port:

    The outward journey places the ship north-east then south-east of port.

    East component of leg 1: 40 sin(50°) = 40 × 0.7660 = 30.64 km

    North component of leg 1: 40 cos(50°) = 40 × 0.6428 = 25.71 km

    East component of leg 2: 65 sin(140°) = 65 × 0.6428 = 41.78 km

    North component of leg 2: 65 cos(140°) = 65 × (−0.7660) = −49.79 km (i.e. southward)

    Total East: 30.64 + 41.78 = 72.42 km

    Total North: 25.71 − 49.79 = −24.08 km (i.e. 24.08 km south)

    To return to port, travel west and north: bearing in third quadrant (S to W? No — need to go west and north).

    Return bearing = bearing of (West 72.42, North 24.08) from current position:

    Reference angle = tan−1(72.42 ÷ 24.08) = tan−1(3.006) = 71.6° west of north

    Return bearing = 360° − 71.6° = 288.4° (or N 71.6° W)

  14. Q14 — Two observers and tower height

    Problem Solving

    Two observers A and B are 300 m apart on level ground, both on the same side of a vertical tower. Observer A sees the top of the tower at an elevation of 35° and observer B (closer to the tower) sees it at 52°.

    (a) Set up two equations linking tower height h and horizontal distance x from A to the base of the tower.

    (b) Find the height of the tower.

    (c) Find the horizontal distance from A to the base of the tower.

    (a) Setting up equations:

    Let h = height of tower, x = horizontal distance from A to base of tower.

    Observer B is between A and the tower, so B is (x − 300) from the base.

    Wait: B is closer to the tower and has a larger elevation angle, so B is between A and the tower.

    Distance from B to base = x − 300 (assuming x > 300).

    From A: tan(35°) = h/x → h = x tan(35°)    ...(1)

    From B: tan(52°) = h/(x − 300) → h = (x − 300) tan(52°)    ...(2)

    (b) Finding tower height:

    Set (1) = (2):

    x tan(35°) = (x − 300) tan(52°)

    x × 0.7002 = (x − 300) × 1.2799

    0.7002x = 1.2799x − 383.97

    383.97 = 1.2799x − 0.7002x = 0.5797x

    x = 383.97 ÷ 0.5797 = 662.3 m

    h = 662.3 × tan(35°) = 662.3 × 0.7002 = 463.7 m

    (c) Horizontal distance from A to base:

    x = 662.3 m

  15. Q15 — Land parcel: angles, area and value

    Problem Solving

    A triangular land parcel has three sides measured as 95 m, 130 m and 160 m.

    (a) Find all three angles using the cosine rule.

    (b) Find the area of the land parcel.

    (c) If land in this area is valued at $1850 per 100 m², find the total value of the parcel.

    Let a = 95 m, b = 130 m, c = 160 m.

    (a) Finding all three angles:

    Angle A (opposite a = 95):

    cos A = (b² + c² − a²) ÷ (2bc)

    cos A = (130² + 160² − 95²) ÷ (2 × 130 × 160)

    cos A = (16 900 + 25 600 − 9025) ÷ 41 600

    cos A = 33 475 ÷ 41 600 = 0.8047

    A = cos−1(0.8047) = 36.5°

    Angle B (opposite b = 130):

    cos B = (a² + c² − b²) ÷ (2ac)

    cos B = (9025 + 25 600 − 16 900) ÷ (2 × 95 × 160)

    cos B = 17 725 ÷ 30 400 = 0.5831

    B = cos−1(0.5831) = 54.3°

    Angle C (opposite c = 160):

    C = 180 − 36.5 − 54.3 = 89.2°

    Check: 36.5 + 54.3 + 89.2 = 180° ✓

    (b) Area of the parcel:

    Area = ½ × a × b × sin C = ½ × 95 × 130 × sin(89.2°)

    = ½ × 12 350 × 0.9999 = 6174.9 m²

    (c) Total value:

    Value = Area ÷ 100 × $1850

    = 6174.9 ÷ 100 × $1850

    = 61.749 × $1850

    = $114 236 (to nearest dollar)