Step 1: Set up the two right triangles without introducing extra unknowns for BC.
Let the height of the building be $h$ metres, let $C$ be the foot of the building, $D$ the top of the building, $B$ the nearer point on the ground and $A$ the farther point, where $AB = 24$ m. Let $BC = x$.
From $B$, the angle of elevation to the top is $60^{\circ}$, so in the right triangle $DBC$:
\[ \tan 60^{\circ} = \frac{h}{x} \]
From $A$, the angle of elevation is $30^{\circ}$, and $AC = x + 24$, so in the right triangle $DAC$:
\[ \tan 30^{\circ} = \frac{h}{x + 24} \]
Step 2: Divide the two tangent equations to cancel out $h$ first.
Instead of solving each equation separately for $h$, divide the first equation by the second, since $h$ appears in both:
\[ \frac{\tan 60^{\circ}}{\tan 30^{\circ}} = \frac{h/x}{h/(x+24)} = \frac{x + 24}{x} \]
Substitute the standard values $\tan 60^{\circ} = \sqrt{3}$ and $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$:
\[ \frac{\sqrt{3}}{1/\sqrt{3}} = \sqrt{3} \times \sqrt{3} = 3 \]
So:
\[ \frac{x + 24}{x} = 3 \]
Step 3: Solve this simpler equation for $x$.
\[ x + 24 = 3x \]
\[ 24 = 3x - x \]
\[ 24 = 2x \]
\[ x = 12 \text{ m} \]
This means $BC = 12$ m.
Step 4: Use $x$ to find the height and the distance of $A$.
Go back to the first equation, $\tan 60^{\circ} = h/x$, and substitute $x = 12$:
\[ h = x\tan 60^{\circ} = 12\sqrt{3} \]
Using $\sqrt{3} = 1.73$:
\[ h = 12 \times 1.73 = 20.76 \text{ m} \]
The distance of point $A$ from the base is:
\[ AC = x + 24 = 12 + 24 = 36 \text{ m} \]
Final Answer:
The height of the building is 20.76 m and the distance of point A from the base is 36 m.
\[ \boxed{h = 20.76 \text{ m}, \ AC = 36 \text{ m}} \]