Given: $x = 152^{ ext{o}} 4'$, $y = 24^{ ext{o}}$ - NSC Technical Mathematics - Question 3 - 2023 - Paper 2

For this part, we first calculate:

Given: $x = 152^{\text{o}} 4' = 152.067^{\text{o}}$

Now, let's compute: $\frac{x}{2} = \frac{152.067^{\text{o}}}{2} = 76.0335^{\text{o}}$

Next, we add 80°: $\frac{x}{2} + 80^{\text{o}} = 76.0335^{\text{o}} + 80^{\text{o}} = 156.0335^{\text{o}}$

Now we find the secant: $sec(156.0335^{\text{o}}) = \frac{1}{cos(156.0335^{\text{o}})}$

Substituting in the original equation gives: $\frac{1}{2} sec(156.0335^{\text{o}}) = \frac{1}{2} \cdot \frac{1}{cos(156.0335^{\text{o}})} \approx 2.97$.

To find $cosec(\beta)$, we use the relation:

$cosec(\beta) = \frac{1}{sin(\beta)}$

Substituting the value of $sin(\beta) = -\frac{4}{5}$ leads to:

$cosec(\beta) = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$.

For this calculation, we first compute $tan(\beta)$.

Using the Pythagorean identity, we find: $cos(\beta) = -\frac{4}{5} \Rightarrow \beta = 180^{\text{o}} + \arcsin(-\frac{4}{5})$

This leads to: $tan(\beta) = \frac{sin(\beta)}{cos(\beta)} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}$

So, finally: $tan(\beta) + cos(\beta) = \frac{4}{3} - \frac{3}{4} = \frac{16}{12} - \frac{9}{12} = \frac{7}{12}$.

Given: $cos(x) = -sin(56.7^{\text{o}})$

Using the identity $sin(90^{\text{o}} - \theta)$ gives: $cos(x) = -cos(33.3^{\text{o}})$

This implies:

1. $x = 180^{\text{o}} + 33.3^{\text{o}} = 213.3^{\text{o}}$
2. $x = 360^{\text{o}} - 33.3^{\text{o}} = 326.7^{\text{o}}$

Therefore, we find: $x \in \{ 213.3^{\text{o}}, 326.7^{\text{o}} \}$.