In △ABC, ∠A = 30° and ∠B = 90°. If AC = 8 cm, then its area is

1. $16\sqrt{3}$ cm²

2. 16 cm²

3. $8\sqrt{3}$ cm²

4. $6\sqrt{3}$ cm²

Considering right angled △ABC we get,

$\Rightarrow \text{sin 30°} = \dfrac{BC}{AC} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{BC}{8} \\[1em] \Rightarrow BC = 4 \text{ cm}$

Similarly,

$\Rightarrow \text{cos 30°} = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{AB}{8} \\[1em] \Rightarrow AB = 4\sqrt{3} \text{ cm}$

Area of right angle triangle = A

$\therefore A = \dfrac{1}{2} \times \text{ base} \times \text{ height} \\[1em] \Rightarrow A = \dfrac{1}{2} \times 4\sqrt{3} \times 4 \\[1em] \Rightarrow A = 8\sqrt{3}\text{ cm}^2$

Hence, Option 3 is the correct option.