Goal Angle A6D23F

1. **State the problem:** We have an ice hockey goal with height 1.22 m and width 1.83 m. A player stands at point P, with distances \(|PG|=3.5\) m, \(|PL|=4.9\) m, and \(|PR|=5.3\) m. Point T is 30 cm (0.3 m) below the top center of the crossbar. We need to find \(|GL|\) and then the angle of elevation \(\angle TPCI\). 2. **Identify points and distances:** - The crossbar length \(|LR|=1.83\) m. - The center \(C\) is midpoint of \(LR\), so \(|LC|=|CR|=\frac{1.83}{2}=0.915\) m. - The height of the goal is 1.22 m, so the top of the crossbar is at height 1.22 m. - Point \(T\) is 0.3 m below the top center, so height of \(T\) is \(1.22 - 0.3 = 0.92\) m. 3. **Find \(|GL|\):** - Points \(P, L, R\) and \(G\) lie on the ground plane. - Given \(|PL|=4.9\) m, \(|PR|=5.3\) m, and \(|PG|=3.5\) m. - We want \(|GL|\), the horizontal distance from \(G\) to \(L\). 4. **Use the fact that \(G\) lies on the goal line between \(L\) and \(R\):** - Since \(|LR|=1.83\) m, and \(G\) lies between \(L\) and \(R\), let \(|GL|=x\), so \(|GR|=1.83 - x\). 5. **Apply the Law of Cosines in triangles \(PLG\) and \(PRG\):** - Triangle \(PLG\): \(|PL|^2 = |PG|^2 + |GL|^2 - 2|PG||GL|\cos(\theta)\) - Triangle \(PRG\): \(|PR|^2 = |PG|^2 + |GR|^2 - 2|PG||GR|\cos(\theta)\) - Here, \(\theta\) is the angle at \(P\) between lines to \(G\) and \(L\) or \(R\). 6. **Since \(L, G, R\) are collinear, the angle between \(PG\) and \(PL\) differs from that between \(PG\) and \(PR\) by 180 degrees, so \(\cos(\theta)\) is the same in magnitude but opposite in sign. We can set up equations to solve for \(x\). 7. **Use the distance formula for \(PL\) and \(PR\) with \(x\):** - \(4.9^2 = 3.5^2 + x^2 - 2 \times 3.5 \times x \cos(\theta)\) - \(5.3^2 = 3.5^2 + (1.83 - x)^2 - 2 \times 3.5 \times (1.83 - x) \cos(\theta)\) 8. **Subtract the two equations to eliminate \(\cos(\theta)\):** $$4.9^2 - 5.3^2 = x^2 - (1.83 - x)^2 - 2 \times 3.5 \cos(\theta) (x - (1.83 - x))$$ $$= x^2 - (1.83^2 - 2 \times 1.83 x + x^2) - 2 \times 3.5 \cos(\theta) (2x - 1.83)$$ $$= x^2 - 1.83^2 + 3.66 x - x^2 - 7 \times x \cos(\theta) + 2 \times 3.5 \times 1.83 \cos(\theta)$$ $$= 3.66 x - 1.83^2 - 7 x \cos(\theta) + 12.81 \cos(\theta)$$ 9. **Calculate left side:** $$4.9^2 - 5.3^2 = 24.01 - 28.09 = -4.08$$ 10. **Rearranged equation:** $$-4.08 = 3.66 x - 3.35 - 7 x \cos(\theta) + 12.81 \cos(\theta)$$ 11. **We have two unknowns \(x\) and \(\cos(\theta)\). Use one of the original equations to express \(\cos(\theta)\) in terms of \(x\):** From \(PL\) equation: $$4.9^2 = 3.5^2 + x^2 - 2 \times 3.5 \times x \cos(\theta)$$ $$24.01 = 12.25 + x^2 - 7 x \cos(\theta)$$ $$7 x \cos(\theta) = x^2 + 12.25 - 24.01 = x^2 - 11.76$$ $$\cos(\theta) = \frac{x^2 - 11.76}{7 x}$$ 12. **Substitute \(\cos(\theta)\) into the rearranged equation:** $$-4.08 = 3.66 x - 3.35 - 7 x \times \frac{x^2 - 11.76}{7 x} + 12.81 \times \frac{x^2 - 11.76}{7 x}$$ $$-4.08 = 3.66 x - 3.35 - (x^2 - 11.76) + \frac{12.81 (x^2 - 11.76)}{7 x}$$ 13. **Multiply both sides by \(7 x\) to clear denominator:** $$-4.08 \times 7 x = 7 x (3.66 x - 3.35 - x^2 + 11.76) + 12.81 (x^2 - 11.76)$$ $$-28.56 x = 25.62 x^2 - 23.45 x - 7 x^3 + 82.32 x + 12.81 x^2 - 150.62$$ 14. **Combine like terms:** $$-28.56 x = (25.62 + 12.81) x^2 + (-23.45 + 82.32) x - 7 x^3 - 150.62$$ $$-28.56 x = 38.43 x^2 + 58.87 x - 7 x^3 - 150.62$$ 15. **Bring all terms to one side:** $$0 = -7 x^3 + 38.43 x^2 + 58.87 x + 28.56 x - 150.62$$ $$0 = -7 x^3 + 38.43 x^2 + 87.43 x - 150.62$$ 16. **Multiply both sides by \(-1\) to simplify:** $$0 = 7 x^3 - 38.43 x^2 - 87.43 x + 150.62$$ 17. **Solve cubic equation approximately (numerical methods or trial):** - Try \(x=1.5\): $$7(1.5)^3 - 38.43(1.5)^2 - 87.43(1.5) + 150.62 = 7(3.375) - 38.43(2.25) - 131.15 + 150.62$$ $$= 23.625 - 86.47 - 131.15 + 150.62 = -43.375$$ (too low) - Try \(x=2.5\): $$7(15.625) - 38.43(6.25) - 87.43(2.5) + 150.62 = 109.375 - 240.19 - 218.58 + 150.62 = -198.77$$ (too low) - Try \(x=0.5\): $$7(0.125) - 38.43(0.25) - 87.43(0.5) + 150.62 = 0.875 - 9.61 - 43.72 + 150.62 = 97.16$$ (too high) - Try \(x=1.0\): $$7 - 38.43 - 87.43 + 150.62 = 31.76$$ (too high) - Try \(x=1.3\): $$7(2.197) - 38.43(1.69) - 87.43(1.3) + 150.62 = 15.38 - 64.95 - 113.66 + 150.62 = -12.61$$ (close) - Try \(x=1.2\): $$7(1.728) - 38.43(1.44) - 87.43(1.2) + 150.62 = 12.1 - 55.33 - 104.92 + 150.62 = 2.47$$ (close) - Try \(x=1.25\): $$7(1.95) - 38.43(1.56) - 87.43(1.25) + 150.62 = 13.65 - 59.98 - 109.29 + 150.62 = -4.99$$ 18. **By interpolation, \(x \approx 1.22\) m. So, \(|GL| \approx 1.22\) m.** 19. **Find angle of elevation \(\angle TPCI\):** - The vertical height from \(P\) to \(T\) is \(0.92\) m. - The horizontal distance from \(P\) to \(C\) is \(|PG|=3.5\) m. - Angle of elevation \(\theta = \tan^{-1} \left( \frac{\text{vertical height}}{\text{horizontal distance}} \right) = \tan^{-1} \left( \frac{0.92}{3.5} \right)\). 20. **Calculate angle:** $$\theta = \tan^{-1} \left( \frac{0.92}{3.5} \right) = \tan^{-1} (0.2629) \approx 14.8^\circ$$ **Final answers:** - \(|GL| \approx 1.22\) m - Angle of elevation \(\angle TPCI \approx 14.8^\circ\)