1. **Problem statement:** (a) Find the exact value of $\tan 30^\circ$. (b) Given a right-angled triangle with a $60^\circ$ angle, the side opposite this angle is 6 cm, and the hypotenuse is $x$ cm. Find $x$. 2. **Recall the formula for tangent:** $$\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}$$ For $30^\circ$, the exact value is known from special triangles. 3. **Exact value of $\tan 30^\circ$:** From the $30^\circ$-$60^\circ$-$90^\circ$ triangle ratios, the sides are in ratio $1 : \sqrt{3} : 2$ (opposite $30^\circ$, opposite $60^\circ$, hypotenuse respectively). Thus, $$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ 4. **Finding $x$ in the triangle:** Given: - Angle $60^\circ$ - Opposite side to $60^\circ$ is 6 cm - Hypotenuse is $x$ Using the sine function: $$\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{x}$$ We know $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so: $$\frac{\sqrt{3}}{2} = \frac{6}{x}$$ 5. **Solve for $x$:** Multiply both sides by $x$: $$x \cdot \frac{\sqrt{3}}{2} = 6$$ Divide both sides by $\frac{\sqrt{3}}{2}$: $$x = \frac{6}{\frac{\sqrt{3}}{2}}$$ Show cancellation: $$x = 6 \times \frac{2}{\cancel{\sqrt{3}}} \times \frac{\cancel{\sqrt{3}}}{\sqrt{3}} = \frac{12}{\sqrt{3}}$$ Rationalize the denominator: $$x = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$$ **Final answers:** (a) $\tan 30^\circ = \frac{\sqrt{3}}{3}$ (b) $x = 4\sqrt{3}$ cm