In die diagram hieronder is P(3 ; m) 'n punt op 'n Kartesiese vlak met OP = \sqrt{13} \beta 'n skerphoek - NSC Technical Mathematics - Question 3 - 2021 - Paper 1

We use the following identities:

$sec^2 \beta = 1 + tan^2 \beta$

Substituting the known values:

$sec^2 \beta + tan^2 \beta = (1 + tan^2 \beta) + tan^2 \beta = 1 + 2tan^2 \beta$

Using the triangle's information to find \tan \beta: If we have:

$tan \beta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{13}}{3}$

Thus:

$tan^2 \beta = \left(\frac{\sqrt{13}}{3}\right)^2 = \frac{13}{9}$

Now substituting back:

$sec^2 \beta + tan^2 \beta = 1 + 2 \times \frac{13}{9} = 1 + \frac{26}{9} = \frac{9 + 26}{9} = \frac{35}{9}$

Given that \cos 0 = \frac{1}{2}, we find that:

$\theta = 60^\circ$

Since \tan \alpha = -1:

This occurs at \alpha = 135^\circ \text{ (in the second quadrant)}. Hence the value of \alpha is:

$\alpha = 135^\circ$

Using the values of \alpha and \theta:

$\cos(\alpha - \theta) = \cos(135^\circ - 60^\circ) = \cos(75^\circ)$

Using the known value:

$\cos(75^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}$

We start with:

$2 \tan x + 0.924 = 0$

This simplifies to:

$\tan x = -\frac{0.924}{2} = -0.462$

To find x, we take the arctangent:

$x = \tan^{-1}(-0.462)$

The reference angle is found to be:

$x \approx 24.8^\circ$

Now considering the range \in [0^\circ ; 360^\circ]:

$x = 180^\circ + 24.8^\circ = 204.8^\circ$

and also:

$x = 360^\circ - 24.8^\circ = 335.2^\circ$

Final solutions are:

$x \approx 155^\circ \text{ or } x \approx 335^\circ$