AB represents a vertical netball pole - NSC Mathematics - Question 7 - 2017 - Paper 2

In triangle $ABE$, we can use the tangent function:

$tan(y) = \frac{AB}{BE}$

From this, we find:

$AB = BE \cdot tan(y)$

Now, in triangle $ABC$, we apply the sine function:

$sin(x) = \frac{AB}{AC}$

Rearranging gives us:

$AC = \frac{AB}{sin(x)}$

Now substituting $AB$ from the previous equation:

$AC = \frac{BE \cdot tan(y)}{sin(x)}$

Since $BE = k$, we replace that to find:

$AC = \frac{k \cdot tan(y)}{sin(x)}$

In triangle $ADC$, using the sine rule:

$\angle ADC = 180^\circ - 90^\circ - 2x = 90^\circ - 2x$

Thus, we can state:

$\frac{DC}{AC} = \frac{sin(90^\circ - 2x)}{sin(x)}$

This simplifies to:

$DC = \frac{AC \cdot sin(90^\circ - 2x)}{sin(x)}$

From earlier we know:

$AC = \frac{k \cdot tan(y)}{sin(x)}$

Substituting this in gives:

$DC = \frac{\frac{k \cdot tan(y)}{sin(x)} \cdot cos(2x)}{sin(x)}$

This leads to:

$DC = \frac{k \cdot tan(y) \cdot 2sin(x)cos(x)}{sin^2(x)}$

Upon further simplification, we arrive at:

$DC = 2k \cdot tan(y)$