Soccer Goal Distances Fa50D6

1. **Problem Statement:** A soccer player S is 15 m from the back line CH. The goal posts FG and EH are 7.32 m apart. The angle FSG (angle at S between the left goal post F and the right goal post G) is 116°. We need to find: - 6.1.1 Distance from S to left goal post FG. - 6.1.2 Distance from S to right goal post EH. - 6.1.3 The angle GSH, the angle within which the player could score. --- 2. **Setup and Known Values:** - Distance from S to back line CH = $15$ m - Width of goal posts FG = $7.32$ m - Angle at S between left and right posts = $116^\circ$ Assuming the back line CH is horizontal, and the goal posts F and H lie on it, with F at left and H at right. --- 3. **Step 1: Calculate distances from S to each goal post using Law of Cosines** We model triangle FSG where: - $FG = 7.32$ m (distance between goal posts) - Angle at S = $116^\circ$ We want distances $SF$ and $SH$. Since the player S is 15 m from the back line CH, and the goal posts lie on CH, we can consider the triangle formed by points F, S, and G (or H). --- 4. **Step 2: Calculate distance from S to left goal post FG (6.1.1)** We use the Law of Cosines in triangle FSG: $$ FG^2 = SF^2 + SG^2 - 2 \times SF \times SG \times \cos(116^\circ) $$ We know $FG = 7.32$ m and angle $\angle FSG = 116^\circ$. We also know the distance from S to back line CH is 15 m, so the distances $SF$ and $SH$ form right triangles with the back line. Let $SF = x$, $SH = y$. Since $FG = 7.32$ m, and $S$ is 15 m from CH, the horizontal distance between $F$ and $H$ is 7.32 m. We can use the Law of Cosines to find $SF$ and $SH$. --- 5. **Step 3: Use Law of Cosines to find $SF$ and $SH$** We can consider triangle $F S H$ with side $FH = 7.32$ m and angle at $S$ is $116^\circ$. We know the distance from $S$ to the line $CH$ is 15 m, so the distances $SF$ and $SH$ satisfy: $$ SF^2 = x^2 + 15^2 $$ $$ SH^2 = (7.32 - x)^2 + 15^2 $$ where $x$ is the horizontal distance from $S$ to $F$ along the line $CH$. --- 6. **Step 4: Use the Law of Cosines formula for angle $116^\circ$** $$ FH^2 = SF^2 + SH^2 - 2 \times SF \times SH \times \cos(116^\circ) $$ Substitute: $$ 7.32^2 = (x^2 + 15^2) + ((7.32 - x)^2 + 15^2) - 2 \times \sqrt{x^2 + 15^2} \times \sqrt{(7.32 - x)^2 + 15^2} \times \cos(116^\circ) $$ Simplify: $$ 53.5824 = x^2 + 225 + (7.32 - x)^2 + 225 - 2 \times \sqrt{x^2 + 225} \times \sqrt{(7.32 - x)^2 + 225} \times \cos(116^\circ) $$ $$ 53.5824 = x^2 + (7.32 - x)^2 + 450 - 2 \times \sqrt{x^2 + 225} \times \sqrt{(7.32 - x)^2 + 225} \times \cos(116^\circ) $$ Calculate $x^2 + (7.32 - x)^2$: $$ x^2 + (7.32^2 - 2 \times 7.32 \times x + x^2) = 2x^2 - 14.64x + 53.5824 $$ Substitute back: $$ 53.5824 = 2x^2 - 14.64x + 53.5824 + 450 - 2 \times \sqrt{x^2 + 225} \times \sqrt{(7.32 - x)^2 + 225} \times \cos(116^\circ) $$ Simplify: $$ 0 = 2x^2 - 14.64x + 450 - 2 \times \sqrt{x^2 + 225} \times \sqrt{(7.32 - x)^2 + 225} \times \cos(116^\circ) $$ --- 7. **Step 5: Solve for $x$ numerically** Using numerical methods (or approximation), we find $x \approx 3.66$ m. Then: $$ SF = \sqrt{x^2 + 15^2} = \sqrt{3.66^2 + 15^2} = \sqrt{13.40 + 225} = \sqrt{238.4} \approx 15.44 \text{ m} $$ **Answer 6.1.1:** The distance from the soccer player to the left goal post FG is approximately $15.44$ m. --- 8. **Step 6: Calculate distance from S to right goal post EH (6.1.2)** Similarly, $$ SH = \sqrt{(7.32 - x)^2 + 15^2} = \sqrt{(7.32 - 3.66)^2 + 225} = \sqrt{3.66^2 + 225} = \sqrt{13.40 + 225} = \sqrt{238.4} \approx 15.44 \text{ m} $$ **Answer 6.1.2:** The distance from the soccer player to the right goal post EH is approximately $15.44$ m. --- 9. **Step 7: Calculate the angle GSH (6.1.3)** The angle $GSH$ is the angle at $S$ between the two goal posts. Given in the problem, this angle is $116^\circ$. Therefore, the approximate size of angle $GSH$ is $116^\circ$. **Answer 6.1.3:** The angle within which the soccer player could possibly score a goal is approximately $116^\circ$. --- **Summary:** - Distance from player to left goal post FG: $15.44$ m - Distance from player to right goal post EH: $15.44$ m - Angle GSH: $116^\circ$