1. **Problem statement:** Calculate the perimeter of a right-angled triangle pond where the hypotenuse is 1.46 m and one acute angle is 73°. 2. **Formula and rules:** In a right triangle, the sides relate by the Pythagorean theorem: $$c^2 = a^2 + b^2$$ where $c$ is the hypotenuse. Also, trigonometric ratios relate sides and angles: - $$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$ - $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ 3. **Identify sides:** Let the angle $73^\circ$ be at vertex A, hypotenuse $c=1.46$ m. - Opposite side to $73^\circ$ is $a$. - Adjacent side to $73^\circ$ is $b$. 4. **Calculate sides:** $$a = c \times \sin(73^\circ) = 1.46 \times \sin(73^\circ)$$ $$b = c \times \cos(73^\circ) = 1.46 \times \cos(73^\circ)$$ Calculate values: $$a = 1.46 \times 0.9563 = 1.3952$$ $$b = 1.46 \times 0.2924 = 0.4269$$ 5. **Calculate perimeter:** $$P = a + b + c = 1.3952 + 0.4269 + 1.46 = 3.2821$$ 6. **Round to 1 decimal place:** $$P \approx 3.3$$ m **Final answer:** The perimeter of the pond is **3.3 m**.