Gegewe Q = 42° en P = 71° Bepaal: 3.1.1 cot(P − Q) 3.1.2 cos Q sec P Gegewe 3sec β − 5 = 0 en β ∈ [90° ; 360°] Bepaal sin² β − cos² β met behulp van 'n diagram - NSC Technical Mathematics - Question 3 - 2022 - Paper 2

To calculate cos Q:

$cos(42°) \approx 0.7431.$

To find sec P, we use the relation:

$sec(P) = \frac{1}{cos(P)}$

Now substituting for P:

$sec(71°) = \frac{1}{cos(71°)} \approx 1 / 0.3204 \approx 3.12.$

To solve for β, we first isolate sec β:

$3sec(β) = 5$

Thus:

$sec(β) = \frac{5}{3}$

We know that (sec = \frac{1}{cos(β)}), therefore:

$cos(β) = \frac{3}{5}.$

Using the diagram for the triangle:

$y = \sqrt{5^2 - 3^2} = \sqrt{16} = 4$

Finally, we determine:

$sin^2(β) - cos^2(β) = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.$

We set:

$cos(2x) = tan(29°).$

Next, we know:

$tan(29°) \approx 0.55431.$

Now we can solve for x. We convert to the angle:

$2x = 360° - 56.34° \Rightarrow 2x = 303.66°$

Thus:

$x ≈ 151.83°.$

Considering the other solutions for 0° ≤ 2x < 360° would give us:

$x ≈ 28.17°.$