Question

find the range of possible values for x.
the range is \\(\square < x < \square\\).
(simplify your answers.)

Explanation:

Step1: Apply Hinge Theorem

In a triangle, the larger side is opposite the larger angle. Here, the two triangles share a common side, and the other two sides are 6 and 7 (7 > 6). So the angle opposite 7 (31°) and the angle opposite 6 ((x - 3)°) follow the Hinge Theorem: if one side is longer, its opposite angle is larger, and vice versa. Wait, actually, the sides adjacent to the angles: the side with length 7 is opposite the angle 31°, and the side with length 6 is opposite the angle (x - 3)°. Wait, no, the two triangles have two sides equal (the marked sides) and the included sides 6 and 7. So by the Hinge Theorem, if the included side is longer, the included angle is larger. Wait, the Hinge Theorem states that if two sides of one triangle are congruent to two sides of another triangle, but the third side is longer, then the included angle is larger. Here, the two triangles have the two equal sides (the marked ones) and the third sides 6 and 7. So the triangle with third side 7 has included angle 31°, and the triangle with third side 6 has included angle (x - 3)°. Since 7 > 6, the included angle opposite (or related) should be: Wait, actually, the Hinge Theorem: if \( AB = DE \), \( AC = DF \), and \( BC > EF \), then \( \angle A > \angle D \). So in our case, the two equal sides are the upper sides (marked), the included angles are (x - 3)° and 31°, and the third sides are 6 and 7. Wait, no: the third sides are 6 (opposite (x - 3)°) and 7 (opposite 31°)? Wait, maybe I got it reversed. Let's clarify: the two triangles are formed by the common side (the vertical one). The left triangle has sides: equal side (marked), 6, and the vertical side. The right triangle has sides: equal side (marked), 7, and the vertical side. So the two sides that are equal are the marked ones, and the other two sides are 6 and 7 (and the vertical side is common). So the included angles are (x - 3)° (left) and 31° (right). Now, the side opposite (x - 3)° is 6, and the side opposite 31° is 7. Wait, no: in a triangle, the larger side is opposite the larger angle. So in the right triangle, side 7 is opposite 31°, and in the left triangle, side 6 is opposite (x - 3)°. Wait, but 7 > 6, so the angle opposite 7 (31°) should be larger than the angle opposite 6 ((x - 3)°)? No, that's not right. Wait, no: the left triangle has sides: marked side (let's say length a), vertical side (length b), and base 6. The right triangle has sides: marked side (length a), vertical side (length b), and base 7. So the two triangles have two sides equal (a and b), and the third sides 6 and 7. So by the Hinge Theorem, since 7 > 6, the included angle (the angle between a and b) in the right triangle (31°) is larger than the included angle in the left triangle ((x - 3)°). Wait, no: the Hinge Theorem says that if the third side is longer, the included angle is larger. So if 7 > 6, then the included angle for the right triangle (31°) is larger than the included angle for the left triangle ((x - 3)°). So \( 31^\circ > (x - 3)^\circ \)? Wait, no, that can't be, because then x - 3 < 31, so x < 34, but we also need the angle to be positive. Wait, no, maybe I got the Hinge Theorem reversed. Wait, the Hinge Theorem: 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 than the third side of the second. So converse: if the third side is longer, the included angle is larger. So in our case, the two equal sides are the upper sides (marked), the included angles are (x - 3)° (left) and 31° (right), and the third sides are 6 (left) and 7 (right). So since 7 > 6, the included angle of the right triangle (31°) is larger than the included angle of the left triangle ((x - 3)°). Wait, no: the third side of the right triangle is 7, which is longer than the third side of the left triangle (6). So by Hinge Theorem, the included angle of the right triangle (31°) is larger than the included angle of the left triangle ((x - 3)°). So \( 31 > x - 3 \)? Wait, no, that would mean x - 3 < 31, so x < 34. But also, the angle (x - 3)° must be positive, so x - 3 > 0 ⇒ x > 3. But wait, maybe I have the Hinge Theorem reversed. Wait, let's think again. Suppose we have two triangles: Triangle 1: sides a, b, c (c is the third side), included angle θ. Triangle 2: sides a, b, d (d is the third side), included angle φ. If c > d, then θ > φ. So in our case, Triangle 1 (right) has third side 7, included angle 31°. Triangle 2 (left) has third side 6, included angle (x - 3)°. So since 7 > 6, then 31° > (x - 3)°? No, that would mean θ (31°) > φ (x - 3°), so 31 > x - 3 ⇒ x < 34. But also, the angle (x - 3)° must be greater than 0, so x - 3 > 0 ⇒ x > 3. But wait, maybe the other way: if the included angle is larger, the third side is longer. So if (x - 3)° > 31°, then the third side (6) would be longer than 7, which is not true (6 < 7). So (x - 3)° must be less than 31°, so x - 3 < 31 ⇒ x < 34. But also, the angle (x - 3)° must be positive, so x - 3 > 0 ⇒ x > 3. Wait, but that can't be, because 6 < 7, so the angle opposite 6 (which is (x - 3)°) should be less than the angle opposite 7 (31°), because in a triangle, larger side is opposite larger angle. Wait, yes! In a triangle, the larger side is opposite the larger angle. So in the left triangle, side 6 is opposite angle (x - 3)°, and in the right triangle, side 7 is opposite angle 31°. Since 7 > 6, then 31° > (x - 3)°, so (x - 3)° < 31° ⇒ x < 34. Also, the angle (x - 3)° must be greater than 0°, so x - 3 > 0 ⇒ x > 3. But wait, is there another constraint? Wait, no, because the two triangles share the vertical side and the marked side, so the only angles we are comparing are (x - 3)° and 31°, with their opposite sides 6 and 7. So since 7 > 6, 31° > (x - 3)°, so x - 3 < 31 ⇒ x < 34, and x - 3 > 0 ⇒ x > 3. Wait, but that seems too simple. Wait, maybe I made a mistake. Wait, the 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 in our case, the two congruent sides are the marked sides (let's call them length m) and the vertical side (length n). So Triangle 1: sides m, n, 7, included angle 31°. Triangle 2: sides m, n, 6, included angle (x - 3)°. So by Hinge Theorem, if the included angle of Triangle 1 (31°) is larger than the included angle of Triangle 2 ((x - 3)°), then the third side of Triangle 1 (7) is longer than the third side of Triangle 2 (6), which is true (7 > 6). So that means 31° > (x - 3)°, so x - 3 < 31 ⇒ x < 34. Also, the included angle (x - 3)° must be positive, so x - 3 > 0 ⇒ x > 3. Wait, but is there a lower bound? Wait, no, because the angle just needs to be positive. But wait, maybe the other way: if the included angle of Triangle 2 is larger, then the third side would be longer, but since 6 < 7, the included angle of Triangle 2 must be smaller. So (x - 3)° < 31°, and (x - 3)° > 0°, so 3 < x < 34? But that seems too broad. Wait, maybe I messed up the Hinge Theorem. Let's check the Hinge Theorem again. The Hinge Theorem (SAS Inequality Theorem) states that if two sides of one triangle are congruent to two sides of another triangle, but the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second. So in formula: If \( AB = DE \), \( AC = DF \), and \( \angle A > \angle D \), then \( BC > EF \). So in our case, let's label the triangles: Left triangle: vertices A (top), B (bottom left), C (middle). So sides: AB (marked), AC (vertical), BC = 6. Included angle at A: (x - 3)°. Right triangle: vertices A (top), D (bottom right), C (middle). Sides: AD (marked, so AB = AD), AC (vertical), CD = 7. Included angle at A: 31°. So AB = AD, AC = AC (common side), BC = 6, CD = 7. So by Hinge Theorem, if \( BC < CD \) (6 < 7), then \( \angle BAC < \angle DAC \). So \( (x - 3)^\circ < 31^\circ \), which gives \( x - 3 < 31 \) ⇒ \( x < 34 \). Also, the angle \( (x - 3)^\circ \) must be positive (since it's an angle in a triangle), so \( x - 3 > 0 \) ⇒ \( x > 3 \). Wait, but is there a lower bound? Wait, no, because the angle just needs to be greater than 0. But wait, maybe the problem is that the angle (x - 3)° must be greater than 0, and less than 31°, so 3 < x < 34? But that seems too simple. Wait, maybe I made a mistake. Let's think again. The two triangles have two sides equal (AB = AD, AC = AC), and BC = 6, CD = 7. So by SAS Inequality, since BC < CD, then angle BAC < angle DAC. So angle BAC is (x - 3)°, angle DAC is 31°, so (x - 3) < 31 ⇒ x < 34. Also, angle BAC must be greater than 0, so x - 3 > 0 ⇒ x > 3. So the range is 3 < x < 34? Wait, but that seems too broad. Wait, maybe the problem is that the angle (x - 3)° must be greater than 0, and also, since it's a triangle, the angle must be less than 180°, but since the other angle is 31°, and the sides are 6 and 7, maybe the upper bound is 34, and lower bound is 3. Wait, but let's check with the Hinge Theorem. If x = 3, then (x - 3)° = 0°, which is not a valid angle. If x = 34, then (x - 3)° = 31°, so the two angles would be equal, but since 6 ≠ 7, the angles can't be equal (because in a triangle, equal angles would imply equal opposite sides, but 6 ≠ 7). So the angle (x - 3)° must be less than 31°, so x < 34, and greater than 0, so x > 3. So the range is 3 < x < 34. Wait, but maybe I got the Hinge Theorem reversed. Let's suppose that if BC > CD, then angle BAC > angle DAC. But here BC = 6 < CD = 7, so angle BAC < angle DAC. So that's correct. So (x - 3) < 31 ⇒ x < 34, and (x - 3) > 0 ⇒ x > 3. So the range is 3 < x < 34.

Step2: Solve the inequalities

We have two inequalities:
1. \( x - 3 > 0 \) (angle must be positive)
- Solving: \( x > 3 \)
2. \( x - 3 < 31 \) (from Hinge Theorem, angle is smaller than 31°)
- Solving: \( x < 34 \)

Combining these, we get \( 3 < x < 34 \).

Answer:

\( 3 < x < 34 \)