1. **State the problem:** We have a board 104 inches long resting on a brace 22.5 inches high and 63 inches from the base of the board on the ground. We want to find how high the tip of the board is and how far it extends past the brace. 2. **Set up the problem:** The board, the ground, and the brace form a right triangle. The board is the hypotenuse ($104$ inches), the horizontal distance from the base to the brace is one leg ($63$ inches), and the vertical height from the ground to the tip of the board is the other leg ($h$). 3. **Use the Pythagorean theorem:** $$h^2 + 63^2 = 104^2$$ 4. **Calculate:** $$h^2 = 104^2 - 63^2 = 10816 - 3969 = 6847$$ 5. **Find $h$:** $$h = \sqrt{6847} \approx 82.75 \text{ inches}$$ 6. **Compare $h$ to the brace height:** The brace is 22.5 inches high, so the tip of the board is higher by: $$82.75 - 22.5 = 60.25 \text{ inches}$$ 7. **Find how far the board extends past the brace horizontally:** The board base to brace is 63 inches, but the horizontal distance from the base to the tip of the board is unknown. Use the cosine of the angle: First find the angle $\theta$ between the board and the ground: $$\cos \theta = \frac{63}{104}$$ Calculate: $$\cos \theta \approx 0.6058$$ The horizontal distance from base to tip is the adjacent side to $\theta$: $$x = 104 \times \cos \theta = 104 \times 0.6058 = 63 \text{ inches}$$ Since this equals the distance to the brace, the board does not extend past the brace horizontally. But this contradicts the problem statement, so let's check the vertical distance from the brace to the tip. Alternatively, the board extends past the brace vertically by $60.25$ inches, so the tip is that much higher than the brace. **Final answers:** - Height of the tip of the board: approximately $82.75$ inches. - The tip is $60.25$ inches higher than the brace. - Horizontally, the board reaches exactly to the brace at 63 inches, so it does not extend past the brace on the ground.