A diagram of right angled triangle with base 5m and 60 at the ground

We are given:

• An angle $\theta = 60^\circ$.
• The side adjacent to the $60^\circ$ angle is $5 \text{ m}$.
• The side opposite the $60^\circ$ angle is $x$.
• The hypotenuse (opposite the right angle) is $y$.

The tangent of an angle in a right-angled triangle is the ratio of the length of the opposite side to the length of the adjacent side.

$$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$$

Substitute the given values:

$$\tan(60^\circ) = \frac{x}{5 \text{ m}}$$

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

$$\sqrt{3} = \frac{x}{5 \text{ m}}$$

Multiply both sides by $5 \text{ m}$ to solve for $x$:

$$x = 5\sqrt{3} \text{ m}$$

The value of $x$ is $\boxed{5\sqrt{3} \text{ m}}$.

The cosine of an angle in a right-angled triangle is the ratio of the length of the adjacent side to the length of the hypotenuse.

$$\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$

Substitute the given values:

$$\cos(60^\circ) = \frac{5 \text{ m}}{y}$$

We know that $\cos(60^\circ) = \frac{1}{2}$.

$$\frac{1}{2} = \frac{5 \text{ m}}{y}$$

To solve for $y$, cross-multiply:

$$1 \cdot y = 2 \cdot (5 \text{ m})$$ $$y = 10 \text{ m}$$

The value of $y$ is $\boxed{10 \text{ m}}$.