**A right-angle triangle has an adjacent side 60 mm long and an opposite side 40 mm long. Calculate**

We are given a right-angled triangle with the following information:
- Adjacent side = 60 mm
- Opposite side = 40 mm

To calculate the unknown values of the triangle, we can determine:
1. **The hypotenuse** using the Pythagorean theorem.
2. **The angles** using trigonometric ratios.

### 1. Hypotenuse:
The Pythagorean theorem states:
\[
c^2 = a^2 + b^2
\]
where \( c \) is the hypotenuse, \( a \) is the adjacent side, and \( b \) is the opposite side.

Substituting the values:
\[
c^2 = 60^2 + 40^2
\]
\[
c^2 = 3600 + 1600 = 5200
\]
\[
c = \sqrt{5200} \approx 72.11 \, \text{mm}
\]

### 2. Angles:
- **Angle θ (opposite the 40 mm side):** We can use the tangent ratio:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{40}{60} = \frac{2}{3}
\]
Now, calculate \( \theta \):
\[
\theta = \tan^{-1} \left( \frac{2}{3} \right) \approx 33.69^\circ
\]

- **Other angle (adjacent to the 60 mm side):** Since the angles in a triangle sum to \(180^\circ\), the remaining angle is:
\[
90^\circ - 33.69^\circ = 56.31^\circ
\]

### Summary:
- Hypotenuse: \( 72.11 \, \text{mm} \)
- Angle opposite the 40 mm side: \( 33.69^\circ \)
- Angle opposite the 60 mm side: \( 56.31^\circ \)

Would you like further details or clarifications?

### Related Questions:
1. What happens to the hypotenuse if the opposite side is increased?
2. How do we calculate the sine and cosine of each angle in this triangle?
3. How would the triangle's dimensions change if the adjacent side was doubled?
4. How can we find the area of the triangle?
5. Can we use the Law of Cosines in this problem?

### Tip:
Always verify if the triangle is a right triangle before applying the Pythagorean theorem.

A right-angle triangle has an adjacent side 60 mm long and an opposite side 40 mm long. Calculate

Learn how to calculate the hypotenuse and angles of a right-angle triangle with an adjacent side of 60 mm and an opposite side of 40 mm. Using the Pythagorean theorem and trigonometric ratios, this problem provides a detailed solution for students in Grades 9-11.

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