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

• Solve 3D problems using angles of elevation and depression.
• Apply the sine, cosine, and area rules to non-right angled triangles.
• Determine the area of a triangle given two sides and an included angle.
• Interpret and solve 2D and 3D trigonometric word problems.

Lesson summary

This week, we delve deeper into trigonometry, expanding our knowledge into three-dimensional (3D) problems and further applications. While we have previously worked with trigonometry in two-dimensional settings, many real-world situations require us to understand spatial relationships and calculations in three dimensions. Understanding 3D trigonometry is crucial for careers in architecture, engineering, surveying, navigation (especially important in our large country with extensive road networks and coastal areas), and even game development.

Lesson notes

2.1 3D Trigonometry: Angles of Elevation and Depression, Bearings In 3D trigonometry, we extend our understanding of angles to incorporate spatial orientation. The key is to identify right-angled triangles within the 3D structure and apply the familiar trigonometric ratios (sine, cosine, tangent).

Angle of Elevation: The angle formed between the horizontal line of sight and an object above the horizontal. Imagine looking up at a bird in the sky – the angle from your eye level to the bird is the angle of elevation.

Angle of Depression: The angle formed between the horizontal line of sight and an object below the horizontal. Imagine standing on a cliff and looking down at a boat – the angle from your eye level to the boat is the angle of depression.

Bearings: Bearings are used to describe the direction of one point relative to another. True bearings are measured clockwise from North (000°T). Compass bearings use North, South, East, and West as reference points (e.g., N30°E means 30 degrees East of North).

Example 1: Angle of Elevation A cellphone tower stands on level ground. From a point 80 meters from the base of the tower, the angle of elevation to the top of the tower is 58°. Calculate the height of the tower.

Solution: Diagram: Draw a diagram representing the situation. We have a right-angled triangle with the base being 80m, the height being the tower (what we want to find), and the angle of elevation being 58°.

Identify the trig ratio: We need to relate the opposite side (height) to the adjacent side (distance).

Therefore, we use the tangent function: tan(θ) = Opposite / Adjacent Substitute: tan(58°) = height / 80 Solve for height: height = 80 * tan(58°)

Calculate: height ≈ 128.06 meters Therefore, the height of the cellphone tower is approximately 128.06 meters.

Example 2: Angle of Depression and Bearings An observer on top of a building 30m high sees a car at an angle of depression of 20°. The true bearing of the car from the observer is 125°T. Calculate the horizontal distance from the base of the building to the car and find the bearing of the building from the car.

Solution: Diagram: Draw a diagram. You'll have a right-angled triangle formed by the height of the building, the horizontal distance to the car, and the line of sight.

Horizontal Distance: Use the tangent function (since we know the height and angle of depression): tan(20°) = 30 / distance.

Solving for distance: distance = 30 / tan(20°) ≈ 82.42 m Bearing of the building from the car: The bearing of the building from the car is the reciprocal bearing of the car from the building. To find this, add or subtract 180° from the original bearing. Since 125° tan(32°) = height / 25 Solve: height = 25 * tan(32°) ≈ 15.62 m The height of the flagpole is approximately 15.62 meters. We use the tangent ratio because it directly relates the known adjacent side and the unknown opposite side (height).

Question 2: An airplane is flying at an altitude of 1500 meters. The angle of depression from the airplane to a point on the ground is 28°. What is the horizontal distance from the airplane to the point on the ground?

Solution: Diagram: Draw a right-angled triangle. The altitude is the vertical side (1500m), the angle of depression is 28°, and we're looking for the horizontal distance (adjacent).

Trig Ratio: tan(28°) = Opposite / Adjacent = 1500 / distance Solve: distance = 1500 / tan(28°) ≈ 2820.62 m The horizontal distance is approximately 2820.62 meters. Again, the tangent ratio provides the direct relationship needed to solve for the unknown distance.

Question 3: In triangle ABC, AB = 7 cm, BC = 9 cm, and angle B = 52°. Calculate the area of triangle AB

C. Solution: Identify: We have two sides and an included angle – use the area rule.

Area Rule: Area = (1/2) AB BC * sin B Substitute: Area = (1/2) 7 9 * sin(52°)

Calculate: Area ≈ 24.82 cm² The area of triangle ABC is approximately 24.82 cm². The area rule provides a direct method to calculate the area given these specific parameters. Independent Practice (Questions Only) From a point on the ground, the angle of elevation to the top of a building is 40°. If the point is 20 meters away from the base of the building, find the height of the building. An observer in a lighthouse 25 meters above sea level sights a boat. The angle of depression to the boat is 12°. Calculate the distance of the boat from the base of the lighthouse. Triangle PQR has PQ = 12cm, QR = 8cm and angle PQR = 75°. Calculate the area of triangle PQR. In triangle XYZ, angle X = 35°, angle Y = 80°, and side x = 8 cm. Find the length of side y. A vertical tower is supported by a cable that stretches from the top of the tower to a point on the ground 30m from the base of the tower. The angle of depression from the top of the tower to the point on the ground is 65°. Calculate the height of the tower. In triangle ABC, side a = 10cm, side b = 12cm, and angle C = 50°. Calculate the length of side c.