1. **State the problem:** Lin's walking path forms a triangle with vertices at the gym (G), his house (H), and the lecture hall (L). We know the sides: $GH=615$ ft, $GL=910$ ft, and $HL=651$ ft. We are to find the direction Lin walked from the gym to his house and the type of triangle formed. 2. **Label points and vectors:** Set Lin's house $H$ at the origin $(0,0)$. He walked west 651 ft to the lecture hall $L$, so $L=(-651,0)$. From $L$ he walked northeast 910 ft to the gym $G$, so the vector $LG$ has magnitude 910 and direction northeast (45° from east). Calculate $G$ coordinates: $$G = L + 910(\cos 45^\circ, \sin 45^\circ) = (-651,0) + 910\times \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) = (-651 + 643.54, 643.54) = (-7.46, 643.54)$$ 3. **Find vector from gym to house:** Vector $GH = H - G = (0 + 7.46, 0 - 643.54) = (7.46, -643.54)$. This vector points slightly east and mostly south, so Lin walked **southeast** from the gym to his house. 4. **Type of triangle:** Check angles using Law of Cosines. Use sides: $$a = HL = 651, \quad b = GH = 615, \quad c = GL = 910$$ Check angle at $G$ opposite side $HL$: $$\cos \theta_G = \frac{b^2 + c^2 - a^2}{2bc} = \frac{615^2 + 910^2 - 651^2}{2 \times 615 \times 910} = \frac{378225 + 828100 - 424701}{1119300} = \frac{781624}{1119300} \approx 0.6987$$ $$\theta_G = \cos^{-1}(0.6987) \approx 45.5^\circ$$ Check angle at $H$ opposite side $GL$: $$\cos \theta_H = \frac{a^2 + b^2 - c^2}{2ab} = \frac{651^2 + 615^2 - 910^2}{2 \times 651 \times 615} = \frac{424701 + 378225 - 828100}{800730} = \frac{-253174}{800730} \approx -0.3161$$ $$\theta_H = \cos^{-1}(-0.3161) \approx 108.4^\circ$$ Since one angle is greater than 90°, the triangle is **obtuse**. **Answer:** Lin walked **southeast**, creating an **obtuse** triangle.