1. **State the problem:** We have a right triangle with a base of length 9 and a hypotenuse of length $10 \times \text{pipe size}$. There is an angle of 48 degrees inside the triangle. We need to solve for the "take off" (TO) and find the lengths of the outside arc (OARC), center arc (CARC), and inside arc (IARC). 2. **Identify known values and variables:** - Base (adjacent side) = 9 - Hypotenuse = $10 \times \text{pipe size}$ - Angle = 48° - TO = take off (unknown) 3. **Use trigonometric relationships:** Since the triangle is right angled, we can use cosine to find the adjacent side (base) and sine to find the opposite side (TO). Formula for TO (opposite side): $$\text{TO} = \text{Hypotenuse} \times \sin(\theta)$$ 4. **Calculate TO:** $$\text{TO} = 10 \times \text{pipe size} \times \sin(48^\circ)$$ 5. **Calculate arcs:** Assuming arcs correspond to circular segments with radii related to the pipe size and the triangle sides: - Outside arc radius = $10 \times \text{pipe size} + \text{pipe size} = 11 \times \text{pipe size}$ - Center arc radius = $10 \times \text{pipe size}$ - Inside arc radius = $10 \times \text{pipe size} - \text{pipe size} = 9 \times \text{pipe size}$ Arc length formula: $$\text{Arc length} = \text{Radius} \times \text{Angle in radians}$$ Convert angle to radians: $$48^\circ = \frac{48 \pi}{180} = \frac{4 \pi}{15}$$ 6. **Calculate each arc length:** - Outside arc (OARC): $$\text{OARC} = 11 \times \text{pipe size} \times \frac{4 \pi}{15} = \frac{44 \pi}{15} \times \text{pipe size}$$ - Center arc (CARC): $$\text{CARC} = 10 \times \text{pipe size} \times \frac{4 \pi}{15} = \frac{40 \pi}{15} \times \text{pipe size} = \frac{8 \pi}{3} \times \text{pipe size}$$ - Inside arc (IARC): $$\text{IARC} = 9 \times \text{pipe size} \times \frac{4 \pi}{15} = \frac{36 \pi}{15} \times \text{pipe size} = \frac{12 \pi}{5} \times \text{pipe size}$$ **Final answers:** - $$\text{TO} = 10 \times \text{pipe size} \times \sin(48^\circ)$$ - $$\text{OARC} = \frac{44 \pi}{15} \times \text{pipe size}$$ - $$\text{CARC} = \frac{8 \pi}{3} \times \text{pipe size}$$ - $$\text{IARC} = \frac{12 \pi}{5} \times \text{pipe size}$$