Given \( \cos \theta = \frac{2}{3} \), and \( \theta \) is acute, find \( \sin \theta \).

Use the Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \).

\[ \sin^2 \theta = 1 – \cos^2 \theta = 1 – \left( \frac{2}{3} \right)^2 = 1 – \frac{4}{9} = \frac{5}{9} \]

Since \( \theta \) is acute (\( \sin \theta > 0 \)):

\[ \sin \theta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

Answer: \( \sin \theta = \frac{\sqrt{5}}{3} \)

Find \( \sin 2\theta \).

Use the double-angle formula: \( \sin 2\theta = 2 \sin \theta \cos \theta \).

From (a)(i), \( \sin \theta = \frac{\sqrt{5}}{3} \), and given \( \cos \theta = \frac{2}{3} \).

\[ \sin 2\theta = 2 \cdot \frac{\sqrt{5}}{3} \cdot \frac{2}{3} = \frac{4 \sqrt{5}}{9} \]

Answer: \( \sin 2\theta = \frac{4 \sqrt{5}}{9} \)

Show that \( b = \frac{3a}{4} \).

Method 1: Sine Rule

In triangle \( ABC \), apply the sine rule: \( \frac{b}{\sin \theta} = \frac{a}{\sin 2\theta} \).

From (a): \( \sin \theta = \frac{\sqrt{5}}{3} \), \( \sin 2\theta = \frac{4 \sqrt{5}}{9} \).

\[ \frac{b}{\frac{\sqrt{5}}{3}} = \frac{a}{\frac{4 \sqrt{5}}{9}} \]

\[ b = a \cdot \frac{\frac{\sqrt{5}}{3}}{\frac{4 \sqrt{5}}{9}} = a \cdot \frac{\sqrt{5}}{3} \cdot \frac{9}{4 \sqrt{5}} = a \cdot \frac{9}{12} = \frac{3a}{4} \]

Method 2: Double-Angle Identity

Using \( \sin 2\theta = 2 \sin \theta \cos \theta \):

\[ \frac{b}{\sin \theta} = \frac{a}{2 \sin \theta \cos \theta} \implies b = \frac{a \sin \theta}{2 \sin \theta \cos \theta} = \frac{a}{2 \cos \theta} \]

Substitute \( \cos \theta = \frac{2}{3} \):

\[ b = \frac{a}{2 \cdot \frac{2}{3}} = \frac{a}{\frac{4}{3}} = a \cdot \frac{3}{4} = \frac{3a}{4} \]

Answer: \( b = \frac{3a}{4} \)

Find \( \sin \angle CAD \) in isosceles triangle \( DAC \), where \( \angle CAD \) is the angle at \( A \).

From the diagram, \( \angle CAD = 2\theta \).

From (a)(ii), \( \sin 2\theta = \frac{4 \sqrt{5}}{9} \).

Thus, \( \sin \angle CAD = \sin 2\theta \).

Answer: \( \sin \angle CAD = \frac{4 \sqrt{5}}{9} \)

Find the area of triangle \( DAC \) in terms of \( a \).

Method 1: Using \( \angle CAD \)

Area formula: \( \text{Area} = \frac{1}{2} \cdot \text{AD} \cdot \text{AC} \cdot \sin \angle CAD \).

From the diagram, \( \text{AD} = b \), \( \text{AC} = b \), and from (b), \( b = \frac{3a}{4} \).

From (c), \( \sin \angle CAD = \frac{4 \sqrt{5}}{9} \).

\[ \text{Area} = \frac{1}{2} \cdot \frac{3a}{4} \cdot \frac{3a}{4} \cdot \frac{4 \sqrt{5}}{9} \]

\[ = \frac{1}{2} \cdot \frac{9a^2}{16} \cdot \frac{4 \sqrt{5}}{9} = \frac{9a^2 \cdot 4 \sqrt{5}}{2 \cdot 16 \cdot 9} = \frac{36 \sqrt{5} a^2}{288} = \frac{\sqrt{5} a^2}{8} \]

Method 2: Using \( \angle DCA \) or \( \angle CDA \)

In isosceles triangle \( DAC \), \( \angle DCA = \angle CDA = \theta \), and \( \text{CD} = a \), \( \text{AD} = b = \frac{3a}{4} \).

Area: \( \text{Area} = \frac{1}{2} \cdot \text{CD} \cdot \text{AD} \cdot \sin \angle DCA \).

\[ \text{Area} = \frac{1}{2} \cdot a \cdot \frac{3a}{4} \cdot \sin \theta \]

From (a)(i), \( \sin \theta = \frac{\sqrt{5}}{3} \):

\[ \text{Area} = \frac{1}{2} \cdot a \cdot \frac{3a}{4} \cdot \frac{\sqrt{5}}{3} = \frac{1}{2} \cdot \frac{3a^2}{4} \cdot \frac{\sqrt{5}}{3} = \frac{3 \sqrt{5} a^2}{24} = \frac{\sqrt{5} a^2}{8} \]

Answer: Area of triangle \( DAC = \frac{\sqrt{5} a^2}{8} \)