Mathematics: Applications & Interpretation SL
Geometry and trigonometry — Topic 3
- 1.
A ladder of length 10 m leans against a vertical wall, making an angle of 65° with the horizontal ground. Calculate the height the ladder reaches up the wall.
[2 marks]Marking points
- Uses height = 10 × sin(65°).
- Obtains a height of 9.06 m.
Examiner tip: Sketch the right-angled triangle first: the ladder is the hypotenuse, and the height on the wall is opposite the given angle, so sine is the correct ratio.
- 2.
Marking analysis: A learner attempts the following task: “A ladder of length 10 m leans against a vertical wall, making an angle of 65° with the horizontal ground. Calculate the height the ladder reaches up the wall.” Their response addresses only this point: “Uses height = 10 × sin(65°).” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]Marking points
- Recognises credit for the stated point: Uses height = 10 × sin(65°).
- Identifies the missing requirement: Obtains a height of 9.06 m.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 3.
A building is 50 m tall. An observer stands 120 m from the base of the building on level ground. Calculate the angle of elevation from the observer to the top of the building.
[2 marks]Marking points
- Uses tan(angle) = 50/120.
- Obtains an angle of elevation of 22.6°.
Examiner tip: The angle of elevation is measured from the horizontal upward to the line of sight — always identify the horizontal and vertical legs before choosing tan, sin, or cos.
- 4.
Marking analysis: A learner attempts the following task: “A building is 50 m tall. An observer stands 120 m from the base of the building on level ground. Calculate the angle of elevation from the observer to the top of the building.” Their response addresses only this point: “Uses tan(angle) = 50/120.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]Marking points
- Recognises credit for the stated point: Uses tan(angle) = 50/120.
- Identifies the missing requirement: Obtains an angle of elevation of 22.6°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 5.
In triangle ABC, angle A = 40°, angle B = 65°, and side a = 12 cm. Use the sine rule to find the length of side b.
[3 marks]Marking points
- Writes the sine rule a/sin(A) = b/sin(B).
- Rearranges to b = 12 × sin(65°)/sin(40°).
- Obtains b ≈ 16.9 cm.
Examiner tip: The sine rule pairs each side with the angle directly opposite it — double-check you have matched sides and angles correctly before substituting.
- 6.
Marking analysis: A learner attempts the following task: “In triangle ABC, angle A = 40°, angle B = 65°, and side a = 12 cm. Use the sine rule to find the length of side b.” Their response addresses only this point: “Writes the sine rule a/sin(A) = b/sin(B).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]Marking points
- Recognises credit for the stated point: Writes the sine rule a/sin(A) = b/sin(B).
- Identifies the missing requirement: Rearranges to b = 12 × sin(65°)/sin(40°).
- Identifies the missing requirement: Obtains b ≈ 16.9 cm.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 7.
In triangle ABC, b = 8 cm, c = 10 cm, and the angle between them A = 55°. Use the cosine rule to find the length of side a.
[3 marks]Marking points
- Writes the cosine rule a² = b² + c² − 2bc·cos(A).
- Substitutes a² = 8² + 10² − 2(8)(10)cos(55°).
- Obtains a ≈ 8.50 cm.
Examiner tip: Use the cosine rule when you know two sides and the included angle between them (SAS) — the sine rule cannot be started directly from this information.
- 8.
Marking analysis: A learner attempts the following task: “In triangle ABC, b = 8 cm, c = 10 cm, and the angle between them A = 55°. Use the cosine rule to find the length of side a.” Their response addresses only this point: “Writes the cosine rule a² = b² + c² − 2bc·cos(A).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]Marking points
- Recognises credit for the stated point: Writes the cosine rule a² = b² + c² − 2bc·cos(A).
- Identifies the missing requirement: Substitutes a² = 8² + 10² − 2(8)(10)cos(55°).
- Identifies the missing requirement: Obtains a ≈ 8.50 cm.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 9.
In triangle ABC, a = 7 cm, b = 9 cm, and c = 12 cm. Use the cosine rule to find the size of angle C, the angle opposite the longest side.
[3 marks]Marking points
- Rearranges the cosine rule to cos(C) = (a² + b² − c²)/(2ab).
- Substitutes to obtain cos(C) = (49 + 81 − 144)/126 ≈ −0.111.
- Obtains C ≈ 96.4°, correctly noting the angle is obtuse since cos(C) is negative.
Examiner tip: A negative cosine value always signals an obtuse angle — this is a useful check that you have not mislabelled the triangle's longest side.
- 10.
Marking analysis: A learner attempts the following task: “In triangle ABC, a = 7 cm, b = 9 cm, and c = 12 cm. Use the cosine rule to find the size of angle C, the angle opposite the longest side.” Their response addresses only this point: “Rearranges the cosine rule to cos(C) = (a² + b² − c²)/(2ab).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]Marking points
- Recognises credit for the stated point: Rearranges the cosine rule to cos(C) = (a² + b² − c²)/(2ab).
- Identifies the missing requirement: Substitutes to obtain cos(C) = (49 + 81 − 144)/126 ≈ −0.111.
- Identifies the missing requirement: Obtains C ≈ 96.4°, correctly noting the angle is obtuse since cos(C) is negative.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 11.
Triangle ABC has sides a = 10 cm and b = 14 cm, with the included angle C = 38°. Calculate the area of the triangle.
[2 marks]Marking points
- Uses the area formula Area = (1/2)ab·sin(C).
- Obtains an area of 43.1 cm².
Examiner tip: This area formula only needs two sides and the included angle between them — no height measurement is required.
- 12.
Marking analysis: A learner attempts the following task: “Triangle ABC has sides a = 10 cm and b = 14 cm, with the included angle C = 38°. Calculate the area of the triangle.” Their response addresses only this point: “Uses the area formula Area = (1/2)ab·sin(C).” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]Marking points
- Recognises credit for the stated point: Uses the area formula Area = (1/2)ab·sin(C).
- Identifies the missing requirement: Obtains an area of 43.1 cm².
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 13.
A ship sails 50 km from port A on a bearing of 070° to reach point B, then sails 70 km on a bearing of 140° to reach point C. Calculate the straight-line distance from A to C.
[4 marks]Marking points
- Recognises that the interior angle at B between BA and BC is 180° − (140° − 70°) = 110°.
- Applies the cosine rule: AC² = 50² + 70² − 2(50)(70)cos(110°).
- Evaluates the right-hand side correctly.
- Obtains AC ≈ 99.0 km.
Examiner tip: Draw a diagram marking north lines at each turning point first — the interior angle for the cosine rule is rarely just the difference between the two bearings.
- 14.
Marking analysis: A learner attempts the following task: “A ship sails 50 km from port A on a bearing of 070° to reach point B, then sails 70 km on a bearing of 140° to reach point C. Calculate the straight-line distance from A to C.” Their response addresses only this point: “Recognises that the interior angle at B between BA and BC is 180° − (140° − 70°) = 110°.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: Recognises that the interior angle at B between BA and BC is 180° − (140° − 70°) = 110°.
- Identifies the missing requirement: Applies the cosine rule: AC² = 50² + 70² − 2(50)(70)cos(110°).
- Identifies the missing requirement: Evaluates the right-hand side correctly.
- Identifies the missing requirement: Obtains AC ≈ 99.0 km.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 15.
A boat sails from point A to point B, travelling 40 km east and 30 km north. Calculate the bearing of B from A.
[3 marks]Marking points
- Sketches the displacement and identifies the angle from north as arctan(east/north).
- Calculates arctan(40/30) ≈ 53.1°.
- States the bearing as 053°, measured clockwise from north.
Examiner tip: Bearings are always given as three digits measured clockwise from north — convert your triangle angle to this form at the very end.
- 16.
Marking analysis: A learner attempts the following task: “A boat sails from point A to point B, travelling 40 km east and 30 km north. Calculate the bearing of B from A.” Their response addresses only this point: “Sketches the displacement and identifies the angle from north as arctan(east/north).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]Marking points
- Recognises credit for the stated point: Sketches the displacement and identifies the angle from north as arctan(east/north).
- Identifies the missing requirement: Calculates arctan(40/30) ≈ 53.1°.
- Identifies the missing requirement: States the bearing as 053°, measured clockwise from north.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 17.
A cone has base radius 6 cm and height 8 cm. (a) Find the slant height of the cone. (b) Find the angle between the slant and the base.
[4 marks]Marking points
- Uses Pythagoras' theorem: slant² = 6² + 8².
- Obtains a slant height of 10 cm.
- Uses tan(angle) = 8/6 to find the angle between the slant and the base.
- Obtains an angle of 53.1°.
Examiner tip: The radius, height and slant height of a cone always form a right-angled triangle, with the slant height as the hypotenuse.
- 18.
Marking analysis: A learner attempts the following task: “A cone has base radius 6 cm and height 8 cm. (a) Find the slant height of the cone. (b) Find the angle between the slant and the base.” Their response addresses only this point: “Uses Pythagoras' theorem: slant² = 6² + 8².” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: Uses Pythagoras' theorem: slant² = 6² + 8².
- Identifies the missing requirement: Obtains a slant height of 10 cm.
- Identifies the missing requirement: Uses tan(angle) = 8/6 to find the angle between the slant and the base.
- Identifies the missing requirement: Obtains an angle of 53.1°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 19.
A cuboid has length 5 cm, width 7 cm, and height 9 cm. (a) Find the length of the space diagonal of the cuboid. (b) Find the angle this diagonal makes with the base.
[4 marks]Marking points
- Finds the base diagonal first: √(5² + 7²) ≈ 8.60 cm.
- Finds the space diagonal using √(5² + 7² + 9²) ≈ 12.45 cm.
- Uses tan(angle) = 9/8.60 with the base diagonal and the height.
- Obtains an angle of approximately 46.3°.
Examiner tip: Finding a 3D angle always reduces to a 2D right-angled triangle — identify the vertical height and the correct horizontal diagonal before applying trigonometry.
- 20.
Marking analysis: A learner attempts the following task: “A cuboid has length 5 cm, width 7 cm, and height 9 cm. (a) Find the length of the space diagonal of the cuboid. (b) Find the angle this diagonal makes with the base.” Their response addresses only this point: “Finds the base diagonal first: √(5² + 7²) ≈ 8.60 cm.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: Finds the base diagonal first: √(5² + 7²) ≈ 8.60 cm.
- Identifies the missing requirement: Finds the space diagonal using √(5² + 7² + 9²) ≈ 12.45 cm.
- Identifies the missing requirement: Uses tan(angle) = 9/8.60 with the base diagonal and the height.
- Identifies the missing requirement: Obtains an angle of approximately 46.3°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 21.
A circle has radius 15 cm. Find the length of the arc subtended by an angle of 50° at the centre.
[3 marks]Marking points
- Converts 50° to radians: 50 × π/180 ≈ 0.873 radians.
- Uses arc length = r × θ (in radians).
- Obtains an arc length of 13.1 cm.
Examiner tip: The arc length formula s = rθ requires the angle in radians — always convert from degrees first, or set your calculator to radian mode.
- 22.
Marking analysis: A learner attempts the following task: “A circle has radius 15 cm. Find the length of the arc subtended by an angle of 50° at the centre.” Their response addresses only this point: “Converts 50° to radians: 50 × π/180 ≈ 0.873 radians.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]Marking points
- Recognises credit for the stated point: Converts 50° to radians: 50 × π/180 ≈ 0.873 radians.
- Identifies the missing requirement: Uses arc length = r × θ (in radians).
- Identifies the missing requirement: Obtains an arc length of 13.1 cm.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 23.
A circular sector has radius 9 cm and central angle 120°. Find the area of the sector.
[3 marks]Marking points
- Converts 120° to radians: 120 × π/180 ≈ 2.094 radians.
- Uses sector area = (1/2)r²θ (in radians).
- Obtains an area of 84.8 cm².
Examiner tip: Like the arc length formula, the sector area formula requires radians — this is a common place to lose marks by forgetting to convert.
- 24.
Marking analysis: A learner attempts the following task: “A circular sector has radius 9 cm and central angle 120°. Find the area of the sector.” Their response addresses only this point: “Converts 120° to radians: 120 × π/180 ≈ 2.094 radians.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]Marking points
- Recognises credit for the stated point: Converts 120° to radians: 120 × π/180 ≈ 2.094 radians.
- Identifies the missing requirement: Uses sector area = (1/2)r²θ (in radians).
- Identifies the missing requirement: Obtains an area of 84.8 cm².
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 25.
Vectors a = (3, 4) and b = (−1, 2) represent two displacements. (a) Find the resultant vector a + b. (b) Find the magnitude of the resultant and the angle it makes with the positive x-axis.
[4 marks]Marking points
- Adds the vectors component-wise: a + b = (3 + (−1), 4 + 2) = (2, 6).
- Uses magnitude = √(2² + 6²).
- Obtains a magnitude of 6.32.
- Uses angle = arctan(6/2) to obtain an angle of 71.6° from the positive x-axis.
Examiner tip: Add vectors component by component, never by adding their magnitudes directly — the magnitude of a sum is not the sum of the magnitudes except when vectors point the same way.
- 26.
Marking analysis: A learner attempts the following task: “Vectors a = (3, 4) and b = (−1, 2) represent two displacements. (a) Find the resultant vector a + b. (b) Find the magnitude of the resultant and the angle it makes with the positive x-axis.” Their response addresses only this point: “Adds the vectors component-wise: a + b = (3 + (−1), 4 + 2) = (2, 6).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]Marking points
- Recognises credit for the stated point: Adds the vectors component-wise: a + b = (3 + (−1), 4 + 2) = (2, 6).
- Identifies the missing requirement: Uses magnitude = √(2² + 6²).
- Identifies the missing requirement: Obtains a magnitude of 6.32.
- Identifies the missing requirement: Uses angle = arctan(6/2) to obtain an angle of 71.6° from the positive x-axis.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 27.
Triangle PQR is similar to triangle XYZ, with a scale factor of 1.5 from XYZ to PQR. If side XY = 8 cm, find the length of the corresponding side PQ.
[2 marks]Marking points
- Multiplies the given side by the scale factor: PQ = 8 × 1.5.
- Obtains PQ = 12 cm.
Examiner tip: Make sure the scale factor is applied in the correct direction — multiply when scaling up to the larger similar figure, divide when scaling down.
- 28.
Marking analysis: A learner attempts the following task: “Triangle PQR is similar to triangle XYZ, with a scale factor of 1.5 from XYZ to PQR. If side XY = 8 cm, find the length of the corresponding side PQ.” Their response addresses only this point: “Multiplies the given side by the scale factor: PQ = 8 × 1.5.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]Marking points
- Recognises credit for the stated point: Multiplies the given side by the scale factor: PQ = 8 × 1.5.
- Identifies the missing requirement: Obtains PQ = 12 cm.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 29.
From the top of a vertical cliff 45 m high, the angle of depression to a boat at sea is 25°. Calculate the horizontal distance from the base of the cliff to the boat.
[3 marks]Marking points
- Recognises that the angle of depression equals the angle of elevation from the boat, by alternate angles.
- Uses tan(25°) = 45/d, where d is the horizontal distance.
- Obtains d ≈ 96.5 m.
Examiner tip: The angle of depression is always measured from the horizontal at the observer's eye level downward — it is equal to the angle of elevation measured from the object, not from the base of the cliff.
- 30.
Marking analysis: A learner attempts the following task: “From the top of a vertical cliff 45 m high, the angle of depression to a boat at sea is 25°. Calculate the horizontal distance from the base of the cliff to the boat.” Their response addresses only this point: “Recognises that the angle of depression equals the angle of elevation from the boat, by alternate angles.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]Marking points
- Recognises credit for the stated point: Recognises that the angle of depression equals the angle of elevation from the boat, by alternate angles.
- Identifies the missing requirement: Uses tan(25°) = 45/d, where d is the horizontal distance.
- Identifies the missing requirement: Obtains d ≈ 96.5 m.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
- 31.
In triangle ABC, a = 9 cm, b = 11 cm, and angle C = 50°. (a) Use the cosine rule to find side c. (b) Use the sine rule to find angle A. (c) Hence find angle B.
[6 marks]Marking points
- Applies the cosine rule: c² = 9² + 11² − 2(9)(11)cos(50°).
- Obtains c ≈ 8.64 cm.
- Applies the sine rule: sin(A)/9 = sin(50°)/8.64.
- Obtains A ≈ 52.9°.
- Uses the angle sum of a triangle: B = 180° − 50° − A.
- Obtains B ≈ 77.1°.
Examiner tip: Once you know all three sides and one angle, switch to the sine rule for the remaining angles — it is simpler than applying the cosine rule twice.
- 32.
Marking analysis: A learner attempts the following task: “In triangle ABC, a = 9 cm, b = 11 cm, and angle C = 50°. (a) Use the cosine rule to find side c. (b) Use the sine rule to find angle A. (c) Hence find angle B.” Their response addresses only this point: “Applies the cosine rule: c² = 9² + 11² − 2(9)(11)cos(50°).” Evaluate the response against the complete 6-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[6 marks]Marking points
- Recognises credit for the stated point: Applies the cosine rule: c² = 9² + 11² − 2(9)(11)cos(50°).
- Identifies the missing requirement: Obtains c ≈ 8.64 cm.
- Identifies the missing requirement: Applies the sine rule: sin(A)/9 = sin(50°)/8.64.
- Identifies the missing requirement: Obtains A ≈ 52.9°.
- Identifies the missing requirement: Uses the angle sum of a triangle: B = 180° − 50° − A.
- Identifies the missing requirement: Obtains B ≈ 77.1°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.