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

To find β, we can utilize the area formula for triangle ΔACD:

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

Given:

• Area of ΔACD = 3,3648 × 10⁴ m²
• Distance AC = 233.36 m
• Distance CD = 300 m

Thus:

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

Solving for sin β gives:

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

Calculating:

$$\text{sin} β \approx 0.9612$$

Therefore:

$$β = \text{arcsin}(0.9612) \approx 74° ext{ which makes it } 106° ext{ since } β > 90°.$$

To find the distance AD, we can utilize the previously computed lengths.

Using the law of cosines in triangle ACD:

$$AD^2 = AC^2 + CD^2 - 2 \times AC \times CD \times \text{cos}(∠ACD)$$

Where:

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

Plugging in the values gives:

$$AD^2 = (233.36)^2 + (300)^2 - 2 \times (233.36) \times (300) \times \text{cos}(106°)$$

Calculating:

$$AD \approx 427.84 m$$

Thus, the distance AD is approximately 427.84 m.