**Setting up :** Let $AB = h$ = height of the building, $B$ = foot,
$C$ = second position of the car with $BC = 25$ m and $\angle ACB = 60^\circ$,
$D$ = first position with $\angle ADB = 30^\circ$.

---

### (i) Height of the building

**Step 1 — In right $\triangle ABC$, use $\tan 60^\circ$.**
$$\tan 60^\circ = \frac{AB}{BC} = \frac{h}{25}$$

**Step 2 — Substitute $\tan 60^\circ = \sqrt3$.**
$$\sqrt3 = \frac{h}{25} \ \Rightarrow\ h = 25\sqrt3 \text{ m} \approx 43{\cdot}3 \text{ m}$$

$$\boxed{\text{Height} = 25\sqrt3 \text{ m}}$$

---

### (ii) Distance between the two positions of the car

**Step 3 — In right $\triangle ABD$, use $\tan 30^\circ$.**
$$\tan 30^\circ = \frac{AB}{BD} \ \Rightarrow\ \frac{1}{\sqrt3} = \frac{25\sqrt3}{BD}$$

**Step 4 — Solve for BD.**
$$BD = 25\sqrt3 \times \sqrt3 = 75 \text{ m}$$

**Step 5 — Find CD.**
$$CD = BD - BC = 75 - 25 = 50 \text{ m}$$

$$\boxed{\text{Distance between the two positions} = 50 \text{ m}}$$

---

### (iii) (a) Total time to reach the foot of the building

**Step 6 — Find the uniform speed of the car.**
The car covers $CD = 50$ m in $6$ seconds.
$$\text{Speed} = \frac{50}{6} = \frac{25}{3} \text{ m/s}$$

**Step 7 — Find the time to cover the remaining 25 m (from C to B).**
$$t = \frac{25}{25/3} = 25 \times \frac{3}{25} = 3 \text{ seconds}$$

**Step 8 — Total time from the starting point D to B.**
$$T = 6 + 3 = 9 \text{ seconds}$$

$$\boxed{\text{Total time} = 9 \text{ seconds}}$$

---

### (iii) (b) Distance of the observer from the car at $60^\circ$

**Step 9 — Required distance is AC. Use $\sin 60^\circ$ in $\triangle ABC$.**
$$\sin 60^\circ = \frac{AB}{AC} \ \Rightarrow\ \frac{\sqrt3}{2} = \frac{25\sqrt3}{AC}$$

**Step 10 — Solve for AC.**
$$AC = \frac{25\sqrt3 \times 2}{\sqrt3} = 50 \text{ m}$$

*(Check by Pythagoras : $AC = \sqrt{(25\sqrt3)^2 + 25^2} = \sqrt{1875 + 625} = \sqrt{2500} = 50$ m ✔)*

$$\boxed{AC = 50 \text{ m}}$$