Two-Dimensional Problems (Grade 10 NSC Matric Mathematics): Revision Notes

Worked Example 1: Flying a kite

Problem: Mandla flies a kite on a 17 m string at an inclination of 63°. Find the height of the kite above the ground and the horizontal distance between Mandla and his friend Sipho who stands directly below the kite.

Solution: Step 1: Identify the triangle components

• Hypotenuse = 17 m (string length)
• Angle = 63°
• Height = h (opposite side)
• Distance = d (adjacent side)

Step 2: Calculate the height using sine ratio $\sin 63° = \frac{h}{17}$

• $h = 17 \times \sin 63°$
• $h = 17 \times 0.8910...$
• $h ≈ 15.15 \text{ m}$

Step 3: Calculate the distance using cosine ratio

• $\cos 63° = \frac{d}{17}$

• $d = 17 \times \cos 63°$

• $d = 17 \times 0.4540...$

• $d ≈ 7.72 \text{ m}$

Answer: The kite is 15.15 m above the ground, and Mandla and Sipho are 7.72 m apart.

Worked Example 2: Calculating angles in a quadrilateral

Problem: ABCD is a trapezium with AB = 4 cm, CD = 6 cm, BC = 5 cm and AD = 5 cm. Point E on diagonal AC divides the diagonal such that AE = 3 cm and angle BEC = 90°. Find angle ABC.

Solution: Step 1: Find angle ABE using triangle ABE

In triangle ABE: $\sin ABE = \frac{3}{4}$

$ABE = 48.59...° ≈ 48.6°$

Step 2: Use Pythagoras' theorem to find BE

In triangle ABE: $BE^2 = AB^2 - AE^2 = 4^2 - 3^2 = 7$

$BE = \sqrt{7} \text{ cm}$

Step 3: Find angle CBE using triangle CBE

In triangle CBE: $\cos CBE = \frac{\sqrt{7}}{5} = 0.5291...$

$CBE = 58.05...° ≈ 58.1°$

Step 4: Calculate the total angle

$ABC = ABE + CBE = 48.6° + 58.1° = 106.7°$

Answer: Angle ABC = 106.7°

Worked Example 3: Finding building height

Problem: From a point 100 m away from a building, the angle of elevation to the top is 38.7°. Calculate the height of the building.

Solution: Step 1: Identify the components

• Adjacent side = 100 m
• Angle = 38.7°
• Opposite side = h (height)

Step 2: Use the tangent ratio

$\tan 38.7° = \frac{h}{100}$

Step 3: Solve for h

$h = 100 \times \tan 38.7°$

$h = 100 \times 0.8011...$

$h ≈ 80 \text{ m}$

Answer: The building is 80 m tall.

Worked Example 4: Angles of elevation and depression

Problem: A cellphone tower is 200 m away from a block of flats. From point B, the angle of elevation to the top of the tower (E) is 34°, and the angle of depression to the bottom of the tower (C) is 62°. What is the height of the tower?

Solution:

Step 1: Calculate the height above point B (DE)

Using triangle BDE: $\tan 34° = \frac{DE}{200}$

$DE = 200 \times \tan 34° = 200 \times 0.6745 ≈ 135 \text{ m}$

Step 2: Calculate the height below point B (CD)\

Using triangle BCD: $\tan 62° = \frac{CD}{200}$

$CD = 200 \times \tan 62° = 200 \times 1.8807 ≈ 376 \text{ m}$

Step 3: Add both heights

Total height = DE + CD = 135 + 376 = 511 m

Answer: The cellphone tower is 511 m tall.

Worked Example 5: Construction problem

Problem: Mr Nkosi has a 4 m high garage and wants to add a corrugated iron roof. The roof sheet is 5 m long with a 5° angle. How high must he build the wall?

Solution: Step 1: Identify the triangle Triangle ABC is right-angled with:

• Hypotenuse BC = 5 m
• Angle = 5°
• Height AC = ? (what we need)

Step 2: Use sine ratio

• $\sin 5° = \frac{AC}{5}$
• $AC = 5 \times \sin 5° = 5 \times 0.0872 ≈ 0.4 \text{ m}$

Step 3: Calculate wall height

• Wall height BD = Garage height - AC = 4 - 0.4 = 3.6 m

Answer: Mr Nkosi must build his wall 3.6 m high.