Let $AB$ be the building with $AB=15$ m, $C$ the foot of the ladder on the ground and $AC$ the ladder.

<b>(i) Length of the ladder (at $60^{\circ}$)</b>
Step 1: In right $\triangle ABC$, $AB$ is opposite to the angle of $60^{\circ}$ and $AC$ is the hypotenuse.
$$\sin 60^{\circ}=\frac{AB}{AC}$$
Step 2: Substitute $\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}$ and $AB=15$ m.
$$\frac{\sqrt{3}}{2}=\frac{15}{AC} \;\Rightarrow\; AC=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{ m}$$
Step 3: So the ladder is $10\sqrt{3}\approx 17.32$ m long.

<b>(ii) Distance of the foot of the ladder from the building</b>
Step 4: In right $\triangle ABC$,
$$\tan 60^{\circ}=\frac{AB}{BC}$$
Step 5: Substitute $\tan 60^{\circ}=\sqrt{3}$.
$$\sqrt{3}=\frac{15}{BC} \;\Rightarrow\; BC=\frac{15}{\sqrt{3}}=5\sqrt{3}\text{ m}$$
Step 6: So the distance is $5\sqrt{3}\approx 8.66$ m.

<b>(iii)(a) Width of the road (ladder at $30^{\circ}$)</b>
Step 7: <b>Diagram:</b> Draw the vertical building $AB$ of height 15 m on the left and a vertical wall on the right; the horizontal ground $BD$ between them is the road. The ladder $AD$ rests from the top $A$ of the building to the base $D$ of the opposite wall, making $\angle ADB=30^{\circ}$.
Step 8: In right $\triangle ABD$,
$$\tan 30^{\circ}=\frac{AB}{BD}$$
Step 9: Substitute $\tan 30^{\circ}=\dfrac{1}{\sqrt{3}}$.
$$\frac{1}{\sqrt{3}}=\frac{15}{BD} \;\Rightarrow\; BD=15\sqrt{3}\text{ m}$$
Step 10: Width of the road $=15\sqrt{3}\approx 25.98$ m.

<b>OR (iii)(b) Length of the ladder in this case</b>
Step 11: In right $\triangle ABD$,
$$\sin 30^{\circ}=\frac{AB}{AD}$$
Step 12: Substitute $\sin 30^{\circ}=\dfrac{1}{2}$ and $AB=15$ m.
$$\frac{1}{2}=\frac{15}{AD} \;\Rightarrow\; AD=30\text{ m}$$
Hence the ladder is $30$ m long in this case.