Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply trig rules to solve 3D problems.
• Calculate angles of elevation and depression in 3D.
• Determine angles between lines and planes.
• Draw and interpret 3D diagrams.
• Break down complex 3D problems into 2D triangles.

Lesson summary

Trigonometry, which you studied extensively in Grades 10 and 11, is not just about triangles on a flat piece of paper. In the real world, objects exist in three dimensions. This section extends your trigonometric knowledge to solve problems involving 3D space. This skill is crucial in various fields, from architecture and engineering (designing buildings and bridges) to surveying (mapping land) and navigation (plotting courses for ships and airplanes). Think about the Drakensberg mountains – accurately measuring their height and distances between peaks requires 3D trigonometry.

Lesson notes

2.1 Fundamental Concepts Review: Before diving into 3D, let's recap key 2D trigonometric concepts: Trigonometric Ratios (SOH CAH TOA): In a right-angled triangle: sin θ = Opposite / Hypotenuse cos θ = Adjacent / Hypotenuse tan θ = Opposite / Adjacent Sine Rule: In any triangle ABC: a / sin A = b / sin B = c / sin C Cosine Rule: In any triangle ABC: a² = b² + c² - 2bc cos A cos A = (b² + c² - a²) / 2bc 2.2 3D Trigonometry: Extending to Space In 3D, we deal with solids and their spatial relationships. The key is to visualize and extract relevant 2D triangles within the 3D figure.

Angle of Elevation: The angle formed between the horizontal and the line of sight upwards to an object. Imagine standing on the ground and looking up at the top of a building.

Angle of Depression: The angle formed between the horizontal and the line of sight downwards to an object. Imagine standing on top of a building and looking down at a car on the street.

Line Perpendicular to a Plane: A line is perpendicular to a plane if it is perpendicular to every line lying in that plane that passes through the point of intersection.

Angle Between a Line and a Plane: This is the angle between the line and its projection onto the plane. To find this angle, drop a perpendicular from a point on the line to the plane. The angle between the line and the perpendicular line is 90 degrees. The angle between the line and its projection is the angle of interest.

Angle Between Two Planes (Dihedral Angle): To find the angle between two planes, choose a point on the line of intersection of the two planes. From that point, draw a line perpendicular to the line of intersection in each plane. The angle between these two perpendicular lines is the angle between the planes. 2.3 Strategies for Solving 3D Problems: Draw a Clear Diagram: This is essential. Label all given information. Use different colors or dashed lines to distinguish between lines and planes.

Identify Relevant Triangles: Look for right-angled or non-right-angled triangles within the 3D figure. These triangles will contain the angles and sides you need to find.

Apply Trigonometric Rules: Use SOH CAH TOA, the Sine Rule, and the Cosine Rule to solve for unknown sides and angles within the 2D triangles.

Relate Triangles: Often, you'll need to solve one triangle to find information needed to solve another.

Interpret the Results: Make sure your answer makes sense in the context of the problem. Check units and magnitudes. 2.4 Worked Examples Example 1: From point A, the angle of elevation to the top of a vertical tower BC is 60°. From point D, which is 20m horizontally from A and in line with the base of the tower C, the angle of elevation to the top of the tower BC is 30°. Calculate the height of the tower B

C. Solution: Diagram: Sketch a diagram showing points A, D, C, and

B. BC is the vertical tower. A and D are on the same horizontal plane.

Identify Triangles: Two right-angled triangles: ΔABC and ΔDB

C. Let BC = h. In ΔABC: tan 60° = h/AC => AC = h/tan 60° = h/√3 In ΔDBC: tan 30° = h/DC => DC = h/tan 30° = h√3 We know DC = DA + AC, so h√3 = 20 + h/√3 Multiply through by √3: 3h = 20√3 + h 2h = 20√3 h = 10√3 ≈ 17.32 m Therefore, the height of the tower BC is approximately 17.32 meters.

Example 2: A vertical pole PQ stands on horizontal ground. Points R and S are on the ground. The angle of elevation of P from R is x, and the angle of elevation of P from S is y. RS = d, and angle PRQ = angle PRS = 90°. If RS is perpendicular to both PR and PS, find an expression for the height of the pole PQ in terms of x, y, and d.

Solution: Diagram: Draw a diagram showing the pole PQ, points R and S on the ground, and the given angles. Since PRQ and PRS are 90 degrees and RS is perpendicular to both PR and PS, triangle PRS is a right-angled triangle.

Identify Triangles: Two right-angled triangles: ΔPQR and ΔPQ

S. In ΔPQR: PQ = PR * tan x => PR = PQ / tan x In ΔPQS: PQ = PS * tan y => PS = PQ / tan y In ΔPRS: RS² = PR² + PS² (Pythagorean theorem)

Substitute PR and PS: d² = (PQ / tan x)² + (PQ / tan y)² d² = PQ² (1/tan²x + 1/tan²y) d² = PQ² (cot²x + cot²y) PQ² = d² / (cot²x + cot²y) PQ = d / √(cot²x + cot²y) Therefore, the height of the pole PQ is d / √(cot²x + cot²y).

Example 3: Consider a rectangular prism ABCDEFG

H. AB = 4cm, BC = 3cm, and CG = 5cm. Determine the angle between the planes ABFE and BCG

F. Solution: Diagram: Sketch the rectangular prism.

Line of Intersection: The line of intersection of the planes ABFE and BCGF is B

F. Perpendiculars: In plane ABFE, AB is perpendicular to B

F. In plane BCGF, BC is perpendicular to B

F. Angle Between Planes: The angle between the planes is the angle ABC, which is 90 degrees since ABCDEFGH is a rectangular prism.

Therefore, the angle between the planes ABFE and BCGF is 90 degrees. Guided Practice (With Solutions)

Question 1: A cellphone tower stands vertically on flat ground.