AB is a vertical flagpole that is $oxed{ ext{}}rac{ ext{s}}{ ext{ } ext{ } ext{ } ext{ } } ext{m} ext{ } ext{ } ext{ }$ long - NSC Mathematics - Question 7 - 2022 - Paper 2

Using the sine rule in triangle $ACD$, we relate the sides to their opposite angles: $\frac{CD}{sin(135^{\circ}-x)} = \frac{AD}{sin(45^{\circ})}$ Since we found $AD = 3p$, we rearrange: $CD = 3p \frac{sin(135^{\circ} - x)}{sin(45^{\circ})}$ Applying the compound angle identity: $CD = \frac{3p(sin(135^{\circ})cos(x) - cos(135^{\circ})sin(x))}{\frac{\sqrt{2}}{2}}$ Factoring out yields: $CD = \frac{3p\sqrt{2}(cos(x) + sin(x))}{2} = \frac{3p(sin(x) + cos(x))}{\sqrt{2}sin(x)}$

To find the area of triangle $ADC$, we can use the formula: $Area = \frac{1}{2}AD \cdot CD \cdot sin(A)$ Substituting the known values: $Area = \frac{1}{2}(3p)(CD)(sin(45^{\circ}))$ Given $p = 10$: $Area = \frac{1}{2}(30)(\frac{3(10)(sin(110^{\circ}) + cos(110^{\circ}))}{\sqrt{2}sin(110^{\circ})})sin(45^{\circ})$ After calculating: $Area = 143.11 m^2$