Two birds X and Y on the ground and on the same side of a tree (T) tree at an angle of elevation of 45° and 60° respectively. If the dista 6m: (a) illustrate the information in a diagram; (b) calculate, correct to four significant figures, the height of the t

From the diagram, we have two right-angled triangles.
For bird Y:

$$\tan(60^\circ) = \frac{h}{d_Y}$$ $$d_Y = \frac{h}{\tan(60^\circ)}$$

For bird X:

$$\tan(45^\circ) = \frac{h}{d_X}$$ $$d_X = \frac{h}{\tan(45^\circ)}$$

We know that $d_X = d_Y + 6$.
Substitute the expressions for $d_X$ and $d_Y$ from Step 1:

$$\frac{h}{\tan(45^\circ)} = \frac{h}{\tan(60^\circ)} + 6$$

We know $\tan(45^\circ) = 1$ and $\tan(60^\circ) = \sqrt{3}$.

$$\frac{h}{1} = \frac{h}{\sqrt{3}} + 6$$ $$h - \frac{h}{\sqrt{3}} = 6$$

Factor out $h$:

$$h \left(1 - \frac{1}{\sqrt{3}}\right) = 6$$ $$h \left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) = 6$$ $$h = \frac{6\sqrt{3}}{\sqrt{3}-1}$$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, $(\sqrt{3}+1)$:

$$h = \frac{6\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$$ $$h = \frac{6(\sqrt{3} \cdot \sqrt{3} + \sqrt{3} \cdot 1)}{(\sqrt{3})^2 - (1)^2}$$ $$h = \frac{6(3 + \sqrt{3})}{3 - 1}$$ $$h = \frac{6(3 + \sqrt{3})}{2}$$ $$h = 3(3 + \sqrt{3})$$ $$h = 9 + 3\sqrt{3}$$

Using $\sqrt{3} \approx 1.7320508$:

$$h = 9 + 3(1.7320508)$$ $$h = 9 + 5.1961524$$ $$h = 14.1961524 \text{ m}$$

Rounding to four significant figures:

$$h \approx 14.20 \text{ m}$$

The height of the tree is $\boxed{14.20 \text{ m}}$.