Question

3.) use the hinge theorem to solve this problem. you and a friend take different flights in opposite direction but leave from the same airport. your flight starts out 210 miles due south and then turns 70 degrees west and goes another 80 miles. your friend’s flight starts out traveling 80 miles due north and then turns 50 degrees to the east before traveling 210 mile. explain how the hinge theorem would help you find out whether you or your friend’s flight is farthest from the airport. use the compass for reference and provide a sketch.

Explanation:

Step1: Define the Triangles

Let the airport be point \( A \). For your flight: first leg \( AB = 210 \) miles (south), second leg \( BC = 80 \) miles (70° west from south). For your friend’s flight: first leg \( AD = 80 \) miles (north), second leg \( DE = 210 \) miles (50° east from north). So we have two triangles \( \triangle ABC \) and \( \triangle ADE \) with \( AB = DE = 210 \), \( BC = AD = 80 \).

Step2: Find the Included Angles

For your triangle: The angle at \( B \) (between south and 70° west) is \( 180^\circ - 70^\circ = 110^\circ \)? Wait, no—south to west turn: starting south, turning 70° west, so the angle between \( AB \) (south) and \( BC \) (west-southwest) is \( 90^\circ + 70^\circ \)? Wait, no, better: south is 180° direction, west is 270°, so the angle between \( AB \) (210°? No, compass: south is 180° from north. Wait, initial direction: your flight starts south (180°), then turns 70° west, so new direction is 180° + 70° = 250°, so the angle between the two legs (AB and BC) is the angle between south (180°) and 250°, which is 70°? Wait, no, the included angle at \( A \)? Wait, no, the two triangles: \( AB = 210 \), \( BC = 80 \), and \( AD = 80 \), \( DE = 210 \). The included angle for your triangle: at \( A \), the angle between south (AB) and the next leg. Wait, maybe better: the angle between the two sides of length 210 and 80. For your flight, the first leg is south (let's say along negative y-axis), second leg is 70° west from south, so the angle between AB (south) and BC is 70°? Wait, no, when you go south (down y-axis), then turn 70° west (towards negative x-axis), so the angle between AB (down y-axis) and BC (70° from down y-axis towards left) is 70°. For your friend: first leg north (up y-axis), turn 50° east (towards positive x-axis), so angle between AD (up y-axis) and DE is 50°. Wait, no, the included angle at \( A \) for each triangle: for your triangle, the angle between AB (south) and the line from A to B? Wait, no, the two sides are AB (210) and BC (80), with included angle at B? No, the Hinge Theorem (SAS Inequality) states that if two sides of one triangle are congruent to two sides of another triangle, but the included angle is larger, then the third side is longer. So we need to find the included angles between the two sides of equal length. So \( AB = DE = 210 \), \( BC = AD = 80 \). So in \( \triangle ABC \), sides \( AB = 210 \), \( BC = 80 \), included angle \( \angle ABC \). In \( \triangle ADE \), sides \( AD = 80 \), \( DE = 210 \), included angle \( \angle ADE \). Wait, no, \( AB = DE = 210 \), \( BC = AD = 80 \), so the included angles are \( \angle B \) (in \( \triangle ABC \)) and \( \angle D \) (in \( \triangle ADE \)). Now, calculate the included angles:

For your flight: starting south (AB), then turning 70° west, so the angle between AB (south) and BC (the next leg) is \( 90^\circ + 70^\circ = 160^\circ \)? Wait, no, south is straight down, turning 70° west means from south, turn 70° towards west (left), so the angle between the two legs (AB and BC) is \( 180^\circ - 70^\circ = 110^\circ \)? No, maybe I'm overcomplicating. Let's use the compass: north and south are opposite (180° apart), east and west are opposite (180° apart). Your flight: AB is south (length 210), BC is 80 miles at 70° west of south. So the angle between AB (south) and BC is 70° (since turning 70° west from south). Your friend’s flight: AD is north (length 80), DE is 210 miles at 50° east of north. So the angle between AD (north) and DE is 50°. Now, the angle between south (AB) and the line back to A? Wait, no, the included angle for the two triangles: in \( \triangle ABC \), the sides are AB = 210, BC = 80, and in \( \triangle ADE \), sides are AD = 80, DE = 210. The included angle for \( \triangle ABC \) is the angle between AB and BC, which is 180° - 70° = 110°? Wait, no, if you go south (AB), then turn 70° west, the angle between AB (south) and BC (the new direction) is 70°, but the angle at A (the airport) between the two paths? Wait, maybe the included angle is at A. Let's model the airport as the origin. Your path: from A (0,0) go south to B (0, -210), then turn 70° west (towards negative x-axis) and go 80 miles: so the vector from B to C is 80*cos(180° + 70°) in x, 80*sin(180° + 70°) in y? Wait, no, direction: south is (0, -1), turning 70° west means the direction is 180° + 70° = 250° from positive x-axis. So the displacement from B to C is 80*cos(250°) i + 80*sin(250°) j. Then the position of C is (0 + 80*cos(250°), -210 + 80*sin(250°)). For your friend: from A (0,0) go north to D (0, 80), then turn 50° east (direction 50° from positive x-axis) and go 210 miles: displacement from D to E is 210*cos(50°) i + 210*sin(50°) j, so E is (210*cos(50°), 80 + 210*sin(50°)). Now, the distance from A to C and A to E. But using Hinge Theorem: two triangles with two sides equal (AB = DE = 210, BC = AD = 80), and the included angle. The included angle for your triangle (at B? No, at A). Wait, maybe the angle between AB and AD: AB is south (0, -210), AD is north (0, 80), so the angle between AB and AD is 180° (since they're opposite directions). Then the angle between BC and DE: BC is 80 at 250°, DE is 210 at 50°. Wait, this is getting too complex. Let's recall Hinge Theorem: if two sides of one triangle are congruent to two sides of another triangle, but the included angle is larger, then the third side is longer. So we have two triangles: \( \triangle ABC \) (your flight) with sides \( AB = 210 \), \( BC = 80 \), and \( \triangle ADE \) (friend’s flight) with sides \( AD = 80 \), \( DE = 210 \). So \( AB = DE \), \( BC = AD \). The included angle for \( \triangle ABC \) is the angle between \( AB \) and \( BC \), which is 180° - 70° = 110°? Wait, no, when you turn 70° west from south, the angle between the two legs (AB and BC) is 180° - 70° = 110°? For your friend, turning 50° east from north, the angle between AD (north) and DE is 50°, so the angle between the two legs (AD and DE) is 50°. Wait, but the included angle in the Hinge Theorem is between the two equal sides. So in \( \triangle ABC \), sides \( AB = 210 \) and \( BC = 80 \), included angle \( \angle B \). In \( \triangle ADE \), sides \( AD = 80 \) and \( DE = 210 \), included angle \( \angle D \). But \( AB = DE \) and \( BC = AD \), so we can consider the included angles at \( B \) and \( D \). Wait, maybe the angle at \( A \) is the included angle. Let's calculate the angle between the two paths at the airport. Your path: first leg south (180°), second leg 70° west of south (250°), so the angle between the two legs (at A) is 250° - 180° = 70°? No, the angle between the two vectors from A: AB is (0, -210), BC is the vector from B to C, but we need the angle at A between the two paths. Wait, maybe better: the angle between the two sides of length 210 and 80. For your triangle, the two sides are 210 (south) and 80 (west-southwest), the included angle is 180° - 70° = 110°? For your friend, the two sides are 80 (north) and 210 (east-northeast), included angle is 90° - 50° = 40°? No, this is confusing. Wait, the Hinge Theorem says that if two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first is longer. So let's check the included angles:

- Your triangle: sides 210 and 80, included angle. Let's find the angle between the two legs at the vertex where the two sides meet. For your flight, the two legs are 210 (south) and 80 (after turning 70° west), so the angle between them is 180° - 70° = 110°? Wait, no, if you go south, then turn 70° west, the angle between the original direction (south) and the new direction is 70°, so the included angle is 180° - 70° = 110°? For your friend, go north, turn 50° east, so the angle between north and the new direction is 50°, so included angle is 90° - 50° = 40°? No, north to east turn: north is 0°, east is 90°, so turning 50° east from north is 50° from north towards east, so the angle between north (0°) and the new direction (50°) is 50°, so included angle is 50°. Wait, now the two included angles: your triangle has included angle 180° - 70° = 110°? No, maybe the angle between the two sides is 180° - 70° = 110° (since south to west turn: 90° + 70°? No, south is down, west is left, so the angle between down and left-down is 70°, so the angle between the two legs (AB and BC) is 180° - 70° = 110°? Wait, I think I made a mistake. Let's use the linear pair: south and north are opposite (180°), so the angle between your first leg (south) and your friend’s first leg (north) is 180°. Then, your turn is 70° west, friend’s turn is 50° east. So the total included angle for your triangle (at the vertex where the two sides of length 210 and 80 meet) is 180° - 70° + 90°? No, this is too time-consuming. Let's recall: the Hinge Theorem (SAS Inequality) states that if \( \triangle ABC \) and \( \triangle DEF \) have \( AB = DE \), \( BC = EF \), and \( \angle B > \angle E \), then \( AC > DF \). So in our case, let's set \( AB = DE = 210 \), \( BC = AD = 80 \). So \( \triangle ABC \) has sides \( AB = 210 \), \( BC = 80 \), included angle \( \angle B \); \( \triangle ADE \) has sides \( AD = 80 \), \( DE = 210 \), included angle \( \angle D \). Now, \( \angle B \) is the angle between south (AB) and the turn (BC). If you go south, then turn 70° west, the angle between AB (south) and BC is 180° - 70° = 110°? Wait, no, the angle between AB (direction 180°) and BC (direction 180° + 70° = 250°) is 70°, but the internal angle at B? No, the vertex is A (the airport). Wait, I think the correct included angle is at A. Let's calculate the angle at A between the two paths. Your path: from A, go south (vector \( (0, -210) \)), then from B, go 80 miles at 70° west of south (vector \( 80 \cos(250°), 80 \sin(250°) \)). Your friend’s path: from A, go north (vector \( (0, 80) \)), then from D, go 210 miles at 50° east of north (vector \( 210 \cos(50°), 210 \sin(50°) \)). Now, the distance from A to C is \( |C - A| = |B + (BC \text{ vector})| = |(0, -210) + (80 \cos(250°), 80 \sin(250°))| \). The distance from A to E is \( |E - A| = |D + (DE \text{ vector})| = |(0, 80) + (210 \cos(50°), 210 \sin(50°))| \). But using Hinge Theorem: the two triangles have two sides equal (210 and 80), and the included angle. The included angle for your triangle (at A) is the angle between the south direction (AB) and the line AC, but no, the included angle is between the two sides of length 210 and 80. Wait, maybe the included angle is 180° - 70° = 110° for your triangle, and 50° for your friend’s, so since 110° > 50°, the third side (distance from airport) for your flight is longer? Wait, no, maybe I got the angles reversed. Let's check the turn directions: your turn is west (left), friend’s is east (right), so the included angle for your triangle is larger. Thus, by Hinge Theorem, since the included angle is larger, the third side (distance from airport) is longer for your flight? Wait, no, let's recast:

Let \( AB = 210 \), \( BC = 80 \), \( \angle B = 180^\circ - 70^\circ = 110^\circ \) (your triangle).

Let \( AD = 80 \), \( DE = 210 \), \( \angle D = 50^\circ \) (friend’s triangle).

Wait, but \( AB = DE = 210 \), \( BC = AD = 80 \), so by Hinge Theorem, if \( \angle B > \angle D \), then \( AC > AE \). Since \( 110^\circ > 50^\circ \), your distance (AC) is greater than your friend’s (AE).

Step3: Sketch the Triangles

- Draw the compass (N, S, E, W) with airport at center.
- For your flight: draw a vertical line down (south) 210 units (AB), then from B, draw a line 80 units at 70° west of south (BC).
- For your friend’s flight: draw a vertical line up (north) 80 units (AD), then from D, draw a line 210 units at 50° east of north (DE).
- Label the airport as A, your final point as C, friend’s as E.

Answer:

By the Hinge Theorem, the flight with the larger included angle between the two legs (210 miles and 80 miles) will be