AB is a vertical flagpole that is \(\sqrt{5p}\) metres long - NSC Mathematics - Question 7 - 2022 - Paper 2

To find CD, we use the sine rule in triangle ACD. The angle ACD can be expressed as (135^\circ - x). Thus, we apply the sine rule:

$\frac{CD}{sin(135^\circ - x)} = \frac{AD}{sin(\angle ACD)}$

Substituting for AD and substituting: $CD = \frac{3p \cdot sin(135^\circ - x)}{sin(x)}$

Further simplifying using the compound angle identity: $CD = \frac{3p\left(sin(135^\circ)cos(x) - cos(135^\circ)sin(x)\right)}{sin(x)}$

Using special values, where (sin(135^\circ) = \frac{\sqrt{2}}{2}) and (cos(135^\circ) = -\frac{\sqrt{2}}{2}), we have:

$CD = \frac{3p}{2} \cdot \frac{(\sqrt{2}/2)(cos(x) + sin(x))}{sin(x)}$

Therefore, we obtain: $CD = \frac{3p(sin(x) + cos(x))}{\sqrt{2} sin(x)}.$

The area of triangle ADC can be calculated using the area formula:

$Area = \frac{1}{2} (AD)(CD)sin(\angle ADC).$

Substituting the known values:

• AD = (3p = 30)
• Using the expression for CD: $CD = \frac{3p(sin(x) + cos(x))}{\sqrt{2} sin(x)}$

With (p = 10) and substituting values for the angles:

$Area = \frac{1}{2} (30)(\frac{3(10)(sin(110) + cos(110))}{\sqrt{2} sin(110)})sin(45)$

Calculating the final area yields: $Area = \frac{1}{2} \cdot 30 \cdot \frac{3(10)(0.9397 - 0.6428)}{\sqrt{2} \cdot 0.9397} \cdot \frac{\sqrt{2}}{2}$

Upon final calculations, the area is approximately: $Area \approx 143.11 m^2.$