Step 1: Let $PQ$ be the building of height $h$ m, with $Q$ its base.
Let $BQ = x$ m, so $AQ = x + 24$ m.
Step 2: In right $\triangle PQB$ ($60^\circ$) :
$$\tan 60^\circ = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}} \qquad \ldots (i)$$
Step 3: In right $\triangle PQA$ ($30^\circ$) :
$$\tan 30^\circ = \frac{h}{x + 24} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 24}$$
$$\Rightarrow x + 24 = \sqrt{3}h \qquad \ldots (ii)$$
Step 4: Substitute $(i)$ into $(ii)$ :
$$\frac{h}{\sqrt{3}} + 24 = \sqrt{3}h$$
Step 5: Multiply throughout by $\sqrt{3}$ :
$$h + 24\sqrt{3} = 3h \Rightarrow 2h = 24\sqrt{3} \Rightarrow h = 12\sqrt{3}$$
Step 6: $$h = 12 \times 1{\cdot}73 = 20{\cdot}76\ \text{m}$$
Step 7: Distance of $A$ from the base :
$$AQ = \sqrt{3}h = \sqrt{3} \times 12\sqrt{3} = 36\ \text{m}$$
Hence the height of the building $= 12\sqrt{3} \approx 20{\cdot}76$ m and $AQ = 36$ m.