Let's solve the questions one by one.

Question 5.1

Given $\sin \theta = \frac{3}{5}$ and $90^\circ \leq \theta \leq 180^\circ$, determine the value of the following:

5.1.1 Calculate the value of $\cos \theta$.

Since $\theta$ is in the second quadrant, $\cos \theta$ will be negative. We know: $\sin^2 \theta + \cos^2 \theta = 1$ $\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1$ $\frac{9}{25} + \cos^2 \theta = 1$ $\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$ $\cos \theta = -\frac{4}{5}$

5.1.2 Calculate the value of $\tan \theta$.

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$

5.1.3 Calculate the value of $\sec \theta$.

$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

5.1.4 Prove that $\sin^2 \theta + \cos^2 \theta = 1$.

This is the Pythagorean identity, which has already been used in 5.1.1: $\sin^2 \theta + \cos^2 \theta = \left(\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$

5.1.5 Prove that $\tan \theta \cdot \cot \theta = 1$.

$\tan \theta \cdot \cot \theta = \left(\frac{\sin \theta}{\cos \theta}\right) \cdot \left(\frac{\cos \theta}{\sin \theta}\right) = \frac{\sin \theta \cdot \cos \theta}{\cos \theta \cdot \sin \theta} = 1$

Question 5.2

Given $\sin 51^\circ = 0.78$, determine:

5.2.1 $\sin 129^\circ$

Since $\sin(180^\circ - \theta) = \sin \theta$: $\sin 129^\circ = \sin(180^\circ - 51^\circ) = \sin 51^\circ = 0.78$

Question 5.3

In the diagram, $\triangle ABC$ is a right-angled triangle at $C$. Let $D$ be a point on $AB$ such that $CE \perp AB$.

5.3.1 Prove $\triangle CED \sim \triangle ABC$.

Both triangles share $\angle C$, and $\angle CED = \angle CAB$ (since $CE \perp AB$ and by AA similarity criteria). Therefore, $\triangle CED \sim \triangle ABC$.

5.3.2 Write down the ratio of $\sin D$ in $\triangle ABC$.

$\sin D = \frac{CE}{AC}$

5.3.3 $AB = 13$ units, $BC = 5$ units. Calculate $ED$.

First, use the Pythagorean theorem in $\triangle ABC$: $AC = \sqrt{AB^2 - BC^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ units}$ Since $\triangle CED \sim \triangle ABC$: $\frac{ED}{BC} = \frac{CE}{AC} \quad \text{and} \quad \frac{CE}{AC} = \frac{5}{12}$ Thus, $ED = \frac{5}{12} \times 12 = 5$ units.

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Would you like more details or have any questions? Here are 5 related questions you might consider:

1. How can the sine, cosine, and tangent values change depending on the quadrant?
2. Why is $\cos \theta$ negative in the second quadrant?
3. Can the Pythagorean identity be used in other forms of trigonometric problems?
4. What is the significance of proving triangle similarity in geometry?
5. How does triangle similarity help in solving problems related to right-angled triangles?

Tip: When solving trigonometric problems, always visualize the quadrant and angle to determine the correct sign of the trigonometric ratios.