Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Draw and interpret 2D representations of 3D objects.
• Apply trig ratios (sine, cosine, tangent) in 3D space.
• Use the cosine and sine rules in 3D problems.
• Determine the angle between a line and a plane.
• Solve practical problems involving heights and distances.

Lesson summary

This week, we delve into the fascinating world of 3D Trigonometry and its further applications. Building on your Grade 11 Trigonometry knowledge, we will extend trigonometric principles to solve problems involving three-dimensional space. This is not just an abstract mathematical concept; it has very real-world applications. Imagine architects designing buildings in Johannesburg, surveyors mapping the Drakensberg mountains, or engineers planning infrastructure projects like bridges and dams. All of these professions rely heavily on 3D Trigonometry to ensure accuracy, safety, and efficiency.

Lesson notes

2. 1. Introduction to 3D Space 3D Trigonometry deals with problems involving points, lines, and planes in three dimensions. Visualizing these scenarios is crucial. Often, you will be presented with a diagram or will need to create one from a word problem.

Key elements to understand include: Angle of Elevation: The angle formed between the horizontal line of sight and an object above the horizontal. Imagine looking up at the top of a building.

Angle of Depression: The angle formed between the horizontal line of sight and an object below the horizontal. Imagine looking down from a cliff at a boat.

Horizontal Plane: A flat surface that is perfectly level (like the surface of a calm lake).

Vertical Plane: A plane perpendicular to the horizontal plane (like a wall).

Line of Greatest Slope: The line in a plane that makes the largest angle with the horizontal plane. 2.

2. Trigonometric Ratios in 3D The fundamental trigonometric ratios (sine, cosine, tangent) are still applicable in 3D problems, but we need to apply them carefully within the right-angled triangles we identify within the 3D space.

Remember SOH CAH TOA: Sine (sin θ) = Opposite / Hypotenuse Cosine (cos θ) = Adjacent / Hypotenuse Tangent (tan θ) = Opposite / Adjacent We also use the Sine Rule and Cosine Rule extensively for non-right angled triangles.

Sine Rule: a/sin A = b/sin B = c/sin C Cosine Rule: a² = b² + c² - 2bc cos A 2.

3. Identifying Right-Angled Triangles in 3D Diagrams This is arguably the most important skill. Look for lines that are perpendicular to planes. For instance, a vertical tower is perpendicular to the horizontal ground. This creates a right angle which allows us to use trigonometric ratios. 2.

4. Angle Between a Line and a Plane The angle between a line and a plane is defined as the angle between the line and its projection onto the plane.

To find this angle: Identify the line and the plane. Draw a perpendicular from a point on the line to the plane. This point is the foot of the perpendicular. Connect the point on the line to the foot of the perpendicular. The angle between the original line and the line connecting the point to the foot of the perpendicular is the angle between the line and the plane. 2.

5. Worked Examples Example 1: From a point A on the ground, the angle of elevation to the top of a vertical tower BC is 30°. From a point D, 20m further away from the tower on the same horizontal line, the angle of elevation is 20°. Find the height of the tower.

Solution: Draw a diagram: Draw a vertical line BC representing the tower. Mark points A and D on the ground such that A is closer to the tower than D. Label the angles of elevation as 30° at A and 20° at

D. Label AD = 20m. Let BC = h (the height of the tower) and AB = x.

Formulate equations: In triangle ABC: tan 30° = h/x => x = h/tan 30° In triangle DBC: tan 20° = h/(x+20) => x + 20 = h/tan 20° Solve for h: Substitute x = h/tan 30° into the second equation: h/tan 30° + 20 = h/tan 20° 20 = h/tan 20° - h/tan 30° 20 = h(1/tan 20° - 1/tan 30°) h = 20 / (1/tan 20° - 1/tan 30°) h ≈ 27.46 meters Therefore, the height of the tower is approximately 27.46 meters.

Example 2: A flagpole, TP, stands vertically on horizontal ground. Points Q and R are on the ground. TQ = 21m, TR = 17m and angle QTR = 42°. Calculate the height of the flagpole, TP, if the angle of elevation of T from Q is 14°.

Solution: Draw a diagram: Draw a vertical line TP representing the flagpole. Draw points Q and R on the horizontal ground. Connect T, Q, and R to form triangle QT

R. Label the given lengths and angles.

Find angle TQ: Use the fact that angle of elevation of T from Q is 14 degrees. So angle PQT = 90 degrees Calculate PQ: tan(14°) = TP/PQ so PQ = TP/ tan(14°)

Use Cosine Rule to find QR: QR² = TQ² + TR² – 2(TQ)(TR)cos QTR QR² = (21)² + (17)² – 2(21)(17)cos 42° QR² = 441 + 289 - 714(0.7431) QR² = 730 – 530.44 QR² = 199.56 QR = 14.13m Use Sine Rule in Triangle PQR to find angle PRQ: Since TP is perpendicular to the plane PQR, angle TPQ = TPR = 90 degrees. Let angle PQT be alpha = 14. sin(14)/17 = sin(PRQ)/21 sin(PRQ) = 21sin(14)/17 sin(PRQ) = 0.3019 so PRQ = 17.56 Use TP/ TR = tan(angle TPR) since angle TPR is the angle of elevation from R = gamma TP / 17 = tan(gamma) so gamma = arctan(TP/17). We know PQ = 21, angle PQT = 90 and TQ =

2

1. Use sin 14 = TP/TQ, so TP = 21sin(14) = 5.

0

9. Therefore flagpole is roughly 5.09 high.

Example 3: A rectangular box ABCD.EFGH has dimensions AB = 8cm, BC = 6cm, and CF = 5cm. Calculate the angle between the line AG and the plane ABC

D. Solution: Draw a diagram: Draw a rectangular box and label the vertices. Identify the line AG and the plane ABC

D. Identify the projection of AG onto plane ABCD: The projection of AG onto the plane ABCD is A

C. Calculate AC: Using the Pythagorean theorem in rectangle ABCD, AC² = AB² + BC² = 8² + 6² =

1

0

0. Therefore, AC = 10 cm.

Calculate AG: Using the Pythagorean theorem in right-angled triangle ACG, AG² = AC² + CG² = 10² + 5² = 125.