Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

This page supports the lesson note with a companion video and a short classroom-ready summary.

For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.

Performance objectives

• Represent 3D problems using diagrams.
• Apply sine, cosine, and tangent rules in 3D.
• Calculate angles between lines and planes.
• Calculate angles between two planes (dihedral angles).
• Solve realistic, contextual 3D problems.

Lesson summary

This week, we delve into the fascinating realm of three-dimensional trigonometry. Building upon your existing knowledge of 2D trigonometry (sine, cosine, and tangent rules), we will extend these concepts to solve problems involving objects and angles in three-dimensional space. This is crucial because the world around us is 3D! From engineering bridges to calculating the height of a mountain or positioning a cell phone tower, 3D trigonometry provides essential tools for problem-solving. Ignoring its importance limits our ability to analyze real-world scenarios accurately.

Lesson notes

2.1 Understanding 3D Space and Terminology In 3D trigonometry, we work with objects that have length, width, and height. It's essential to visualize the situation and draw accurate diagrams.

Key terms include: Angle of Elevation: The angle formed between the horizontal line of sight and an object above the observer. 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 observer. Imagine looking down at a boat from the top of a cliff.

Angle between a line and a plane: The angle between the line and its projection onto the plane. This is often the trickiest part. Imagine shining a light directly above a line onto a plane. The shadow the line casts on the plane is its projection. The angle between the original line and its shadow is the angle between the line and the plane.

Angle between two planes (Dihedral Angle): The angle between two planes is found by drawing a line perpendicular to the line of intersection of the two planes on each plane. The angle between those two lines is the dihedral angle. 2.2 Drawing 3D Diagrams Drawing accurate diagrams is paramount.

Here are some tips: Start with a 2D base: Draw a 2D representation of the base of the 3D object.

Add the height: Draw the vertical lines representing the height of the objects.

Connect the vertices: Connect the top points to create the 3D shape.

Use dotted lines for hidden edges: Indicate edges that are behind the object with dotted lines. Clearly label all known sides and angles.

Draw separate triangles (if necessary): Sometimes, it helps to draw the relevant triangles separately for clarity. 2.3 Solving 3D Problems using Trigonometry We can use the familiar trigonometric ratios (sin, cos, tan), the sine rule, and the cosine rule to solve 3D problems. The key is to identify the correct triangles within the 3D diagram and apply the appropriate trigonometric principle.

Example 1: Angle of Elevation A flagpole is standing on level ground. From a point 25 meters away from the base of the flagpole, the angle of elevation to the top of the flagpole is 30 degrees. What is the height of the flagpole?

Solution: Draw a diagram: Draw a right-angled triangle where the flagpole is the vertical side, the distance from the observer is the horizontal side, and the angle of elevation is 30 degrees.

Identify the relevant trigonometric ratio: We have the adjacent side (25m) and we want to find the opposite side (height of the flagpole).

Therefore, we use the tangent ratio.

Apply the tangent ratio: tan(30°) = height / 25 Solve for the height: height = 25 tan(30°) ≈ 25 0.577 ≈ 14.43 meters.

Therefore, the height of the flagpole is approximately 14.43 meters.

Example 2: Angle between a line and a plane A vertical tower TP stands on horizontal ground. Points A and B are on the ground. The angle of elevation of T from A is x and the angle of elevation of T from B is y. Also, AB = d and angle ATB = θ. Show that TP = d / sqrt(cot 2 x + cot 2 y - 2cotxcotycosθ).

Solution: Draw a diagram: Draw the vertical tower T

P. Draw points A and B on the ground. Connect TA and T

B. Identify Right Triangles: Triangles TAP and TBP are right angled at

P. Express AP and BP in terms of TP: From triangle TAP, AP = TPcotx. From triangle TBP, BP = TPcoty.

Apply Cosine Rule in triangle APB: AB 2 = AP 2 + BP 2 - 2AP*BPcosθ. Substitute the expressions for AP, BP and AB. d 2 = (TPcotx) 2 + (TPcoty) 2 - 2(TPcotx)(TPcoty)cosθ Solve for TP: d 2 = TP 2 (cot 2 x + cot 2 y - 2cotxcotycosθ). Thus, TP 2 = d 2 / (cot 2 x + cot 2 y - 2cotxcotycosθ). Taking the square root, TP = d / sqrt(cot 2 x + cot 2 y - 2cotxcotycosθ).

Example 3: Angle Between Two Planes (Dihedral Angle) Consider a roof shaped like an isosceles triangular prism. The base is a rectangle ABCD, and the top edges are AE and BE, where E is the vertex of the isosceles triangle. If AE = BE, find the angle between the plane ABE and the base ABC

D. Let AB= 6m, AD = 4m and AE = 5m.

Solution: Draw a diagram: Draw the isosceles triangular prism.

Find the line of intersection: The line of intersection of planes ABE and ABCD is A

B. Draw perpendiculars to the line of intersection: Let M be the midpoint of AB. Draw EM perpendicular to AB in plane AB

E. Draw DM perpendicular to AB in plane ABC

D. Identify the Dihedral angle: Angle EMD is the angle between the two planes.

Calculate lengths: AM = MB = AB/2 = 3m. Since triangle AME is a right angled triangle (at M), EM = sqrt(AE 2 - AM 2 ) = sqrt(5 2 - 3 2 ) = 4m. Since triangle AMD is a right angled triangle (at D), DM = sqrt(AD 2 + AM 2 ) = sqrt(4 2 + 3 2 ) = 5m.

Consider the triangle EMD: we know EM = 4m and DM = 5m. The length of ED isn't immediately obvious, but we don't need it!

Angle EMD: tan(EMD) = EM/DM = 4/

5. Therefore EMD = arctan(4/5) = 38.66 o Therefore, the angle between the two planes is approximately 38.66 degrees. Guided Practice (With Solutions)

Question 1: A tower stands vertically on level ground.