The problem provides a diagram with two right-angled triangles, TQH and PTH. We are given several angles and one side length. The question itself is cut off, but typically for

$\triangle TQH$ is a right-angled triangle at $H$, with $\angle THQ = 90^\circ$.
We are given $\angle TQH = 45^\circ$.
Let $TH = h$.
Using the tangent function in $\triangle TQH$:

$$\tan(\angle TQH) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{TH}{QH}$$ $$\tan(45^\circ) = \frac{h}{QH}$$

Since $\tan(45^\circ) = 1$:

$$1 = \frac{h}{QH}$$ $$QH = h$$

$\triangle PTH$ is also a right-angled triangle at $H$, with $\angle THQ = 90^\circ$.
We are given $\angle TPQ = 30^\circ$. This is the angle $\angle TPH$.
Using the tangent function in $\triangle PTH$:

$$\tan(\angle TPH) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{TH}{PH}$$ $$\tan(30^\circ) = \frac{h}{PH}$$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$:

$$\frac{1}{\sqrt{3}} = \frac{h}{PH}$$ $$PH = h\sqrt{3}$$

From the diagram, points $P$, $Q$, and $H$ are collinear, with $Q$ between $P$ and $H$.
Therefore, we can write the relationship:

$$PH = PQ + QH$$

We are given $PQ = 5 \text{ cm}$. Substitute the expressions for $PH$ and $QH$ from Step 1 and Step 2:

$$h\sqrt{3} = 5 + h$$

Rearrange the equation to solve for $h$:

$$h\sqrt{3} - h = 5$$

Factor out $h$:

$$h(\sqrt{3} - 1) = 5$$ $$h = \frac{5}{\sqrt{3} - 1}$$

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

$$h = \frac{5}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$$ $$h = \frac{5(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}$$ $$h = \frac{5(\sqrt{3} + 1)}{3 - 1}$$ $$h = \frac{5(\sqrt{3} + 1)}{2}$$

Using $\sqrt{3} \approx 1.732$:

$$h = \frac{5(1.732 + 1)}{2}$$ $$h = \frac{5(2.732)}{2}$$ $$h = \frac{13.66}{2}$$ $$h = \boxed{\text{6.83 cm}}$$