In the diagram below, AB represents a vertical tower - NSC Technical Mathematics - Question 6 - 2019 - Paper 2

The area of triangle ACD can be expressed as:

$$\text{Area} = \frac{1}{2} AC \times CD \times \text{sin}(β)$$

Given:

• Area = 3,3648 x 10^4 m²
• AC = 233.36 m
• CD = 300 m

Setting up the equation:

$$3,3648 x 10^4 = \frac{1}{2} \times 233.36 \times 300 \times \text{sin}(β)$$

Solving for sin(β):

$$\text{sin}(β) = \frac{3,3648 x 10^4 \times 2}{233.36 \times 300}$$

Calculating:

$$\text{sin}(β) \approx 0.96126157$$

Thus,

$$β \approx \text{arcsin}(0.96126157) \approx 106°.$$

To find distance AD:

$$AD = AC + DC - 2 imes AC imes DC imes ext{cos}(∠ACD)$$

Substituting the known values:

• AC = 233.36 m
• DC = 300 m
• ∠ACD = 106°

Calculating:

$$AD = 233.36 + 300 - 2 imes 233.36 imes 300 imes ext{cos}(106°)$$

Calculating:

$$AD \approx 1830.57056 - (233.36 imes 300 imes -0.275637)$$

Simplifying yields:

$$AD \approx 427.84 ext{ m}.$$