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To solve for the perimeter of the given right-angled triangle using trigonometry, we can break it down as follows:

We are given:
- One side (the adjacent side) = 15.6 meters
- One angle = 41°
- The right angle implies that we can use trigonometric ratios to find the other two sides (the opposite and the hypotenuse).

### Step 1: Use the tangent ratio to find the opposite side.
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
\[
\tan(41^\circ) = \frac{\text{opposite}}{15.6}
\]
\[
\text{opposite} = 15.6 \times \tan(41^\circ)
\]
Using a calculator:
\[
\text{opposite} \approx 15.6 \times 0.8693 \approx 13.55 \, \text{m}
\]

### Step 2: Use the Pythagorean theorem to find the hypotenuse.
\[
\text{hypotenuse} = \sqrt{\text{opposite}^2 + \text{adjacent}^2}
\]
\[
\text{hypotenuse} = \sqrt{13.55^2 + 15.6^2}
\]
\[
\text{hypotenuse} \approx \sqrt{183.6 + 243.36} = \sqrt{426.96} \approx 20.66 \, \text{m}
\]

### Step 3: Find the perimeter of the triangle.
\[
\text{Perimeter} = \text{adjacent} + \text{opposite} + \text{hypotenuse}
\]
\[
\text{Perimeter} = 15.6 + 13.55 + 20.66 \approx 49.81 \, \text{m}
\]

Thus, the perimeter of the triangle is approximately **49.8 meters** (rounded to 1 decimal place).

Would you like further details or have any questions?

Here are five related questions:
1. How do the different trigonometric ratios apply to right-angled triangles?
2. What is the significance of using the Pythagorean theorem in such problems?
3. How would this problem change if the given side was the hypotenuse instead?
4. What if the angle provided was the other non-right angle?
5. Can the sine or cosine rules be used for this problem, and how?

**Tip**: Always keep in mind that angles in right-angled triangles can be used to easily calculate missing sides using trigonometric functions like sine, cosine, and tangent.

Using trigonometry, work out the perimeter of the right-angled triangle below. The adjacent side is 15.6 meters, and the angle provided is 41°.

Learn how to find the perimeter of a right-angled triangle with an adjacent side of 15.6 meters and a 41° angle. This problem uses trigonometry (tangent) and the Pythagorean theorem to calculate missing sides and the perimeter. Suitable for students in Grades 9-10.

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