Question

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

Explanation:

Step1: Apply the Hinge Theorem (SAS Inequality Theorem)

In a triangle, if two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is shorter than the third side of the second triangle, then the included angle of the first triangle is smaller than the included angle of the second triangle, and vice versa.

Here, the two triangles share the common side (the altitude), and the other two sides are marked as congruent (the tick marks). The third sides are 6 and 7, with \(6 < 7\).

So, the included angle opposite the shorter side (6) is \((x - 3)^\circ\), and the included angle opposite the longer side (7) is \(31^\circ\). Wait, no—actually, if the third side of the first triangle (with side 6) is shorter than the third side of the second triangle (with side 7), then the included angle of the first triangle should be smaller than the included angle of the second triangle? Wait, no, let's correct that.

Wait, the Hinge Theorem states that if two sides of one triangle are congruent to two sides of another triangle, and the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is larger than the included angle of the second triangle.

In this case, the two triangles have two congruent sides (the ones with the tick marks) and the third sides: one is 6, the other is 7. So, the triangle with the longer third side (7) will have the larger included angle. Wait, no: the side of length 7 is opposite the angle \(31^\circ\), and the side of length 6 is opposite the angle \((x - 3)^\circ\). Wait, actually, the two triangles are formed by the common side (the altitude), the congruent sides (the ones with the tick marks), and the bases (6 and 7). So, the included angles are \((x - 3)^\circ\) (opposite base 6) and \(31^\circ\) (opposite base 7).

Since \(6 < 7\), by the Hinge Theorem, the included angle opposite the shorter side (6) is smaller than the included angle opposite the longer side (7). Wait, no—wait, the Hinge Theorem says that if two sides of one triangle are congruent to two sides of another triangle, and the third side of the first triangle is longer than the third side of the second triangle, then the included angle of the first triangle is larger than the included angle of the second triangle. So, if the third side of the first triangle (let's say Triangle 1: sides with tick marks, base 6) is shorter than the third side of Triangle 2 (sides with tick marks, base 7), then the included angle of Triangle 1 is smaller than the included angle of Triangle 2. Wait, no—actually, the included angle is between the two congruent sides. So, in Triangle 1, the included angle is \((x - 3)^\circ\), and the third side is 6. In Triangle 2, the included angle is \(31^\circ\), and the third side is 7. Since \(6 < 7\), then the included angle of Triangle 1 (\((x - 3)^\circ\)) must be less than the included angle of Triangle 2 (\(31^\circ\))? Wait, no, that's not right. Wait, the Hinge Theorem: if \(AB = DE\), \(AC = DF\), and \(BC > EF\), then \(\angle A > \angle D\). So, in our case, the two congruent sides are like \(AB = DE\) and \(AC = DF\), and the third sides are \(BC = 6\) and \(EF = 7\). So, \(BC < EF\), so \(\angle A < \angle D\). Wait, \(\angle A\) is \((x - 3)^\circ\), and \(\angle D\) is \(31^\circ\)? No, that would mean \((x - 3) < 31\), but also, the angle must be positive. Wait, no, maybe I got the direction wrong. Wait, let's re-express:

Let’s denote the two triangles as \(\triangle ABC\) and \(\triangle DEF\), where \(AB = DE\) (the congruent sides with tick marks), \(AC = DF\) (the common side, the altitude), and \(BC = 6\), \(EF = 7\). Then, the included angles are \(\angle BAC = (x - 3)^\circ\) and \(\angle EDF = 31^\circ\).

Since \(BC = 6 < EF = 7\), by the Hinge Theorem, \(\angle BAC < \angle EDF\)? Wait, no—wait, the Hinge Theorem says that if two sides are congruent, and the third side is longer, then the included angle is larger. So, if \(BC < EF\), then \(\angle BAC < \angle EDF\). Wait, but \(\angle EDF\) is \(31^\circ\), so \((x - 3) < 31\). But also, the angle \((x - 3)^\circ\) must be greater than \(0^\circ\) because it's an angle in a triangle. So, \(x - 3 > 0\) (since angle can't be zero or negative) and \((x - 3) < 31\)? Wait, no, that can't be, because then \(x - 3 < 31\) implies \(x < 34\), and \(x - 3 > 0\) implies \(x > 3\). But that seems too small. Wait, maybe I mixed up the angles. Wait, maybe the included angle for the longer side is larger. Wait, let's check again.

Wait, maybe the two triangles are such that the side with length 7 is opposite the angle \((x - 3)^\circ\), and the side with length 6 is opposite the angle \(31^\circ\)? No, the diagram shows the two angles at the top, with the two sides (with tick marks) meeting at the top vertex, split by the altitude. So, the two triangles are isoceles? No, the two sides with tick marks are congruent, so the two triangles have two congruent sides (the tick-marked sides) and a common side (the altitude). So, the third sides are 6 and 7. So, the included angles are the angles at the top vertex, between the tick-marked sides and the altitude. Wait, maybe the angle with the larger third side (7) has the larger included angle. Wait, if the third side is longer, the included angle is larger. So, since 7 > 6, the included angle opposite the side of length 7 (which is \(31^\circ\))? No, that doesn't make sense. Wait, maybe the included angle for the side of length 7 is \(31^\circ\), and the included angle for the side of length 6 is \((x - 3)^\circ\). So, since 7 > 6, then the included angle for 7 (\(31^\circ\)) should be larger than the included angle for 6 (\((x - 3)^\circ\))? No, that would mean \(31 > (x - 3)\), so \(x - 3 < 31\), so \(x < 34\). But also, the angle \((x - 3)^\circ\) must be greater than \(0^\circ\), so \(x - 3 > 0\), so \(x > 3\). But that seems incorrect. Wait, maybe I have the Hinge Theorem backwards. Let's check the Hinge Theorem again:

Hinge Theorem (SAS Inequality): 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.

Converse of Hinge Theorem (SSS Inequality): If two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer than the third side of the second, then the included angle of the first triangle is larger than the included angle of the second.

Ah! So, the converse is what we need. So, if two sides are congruent (the tick-marked sides and the common altitude), and the third side of the first triangle (6) is shorter than the third side of the second triangle (7), then the included angle of the first triangle (\((x - 3)^\circ\)) is smaller than the included angle of the second triangle (\(31^\circ\))? Wait, no—wait, the third side of the first triangle is 6, the third side of the second is 7. So, 6 < 7, so the included angle of the first triangle is smaller than the included angle of the second triangle. So, \((x - 3) < 31\). But also, the angle must be positive, so \((x - 3) > 0\). Wait, but that would give \(3 < x < 34\), but that seems too broad. Wait, maybe the other way: maybe the included angle for the longer side is larger. Wait, let's think of the two triangles:

Triangle 1: sides with tick marks (let's say length \(a\)), altitude (length \(h\)), and base 6. Included angle: \((x - 3)^\circ\).

Triangle 2: sides with tick marks (length \(a\)), altitude (length \(h\)), and base 7. Included angle: \(31^\circ\).

By the Law of Cosines, in Triangle 1: \(6^2 = a^2 + h^2 - 2ah \cos((x - 3)^\circ)\)

In Triangle 2: \(7^2 = a^2 + h^2 - 2ah \cos(31^\circ)\)

Subtract the first equation from the second:

\(49 - 36 = -2ah \cos(31^\circ) + 2ah \cos((x - 3)^\circ)\)

\(13 = 2ah [\cos((x - 3)^\circ) - \cos(31^\circ)]\)

Since \(2ah\) is positive (lengths are positive), then \(\cos((x - 3)^\circ) - \cos(31^\circ) > 0\)

So, \(\cos((x - 3)^\circ) > \cos(31^\circ)\)

We know that the cosine function is decreasing in the interval \([0^\circ, 180^\circ]\), so if \(\cos A > \cos B\), then \(A < B\) (when \(A, B \in [0^\circ, 180^\circ]\))

Therefore, \((x - 3)^\circ < 31^\circ\)

So, \(x - 3 < 31\) => \(x < 34\)

Also, the angle \((x - 3)^\circ\) must be greater than \(0^\circ\) (since it's an angle in a triangle, it can't be zero or negative), so:

\(x - 3 > 0\) => \(x > 3\)

Wait, but that seems too simple. Wait, but also, the angle must be less than \(180^\circ\), but since it's part of a triangle with two other angles, but in this case, the two triangles are adjacent, so the total angle at the top would be \((x - 3) + 31\), which must be less than \(180^\circ\), but maybe that's not necessary here. Wait, no, the Hinge Theorem gives the relationship between the two angles. Wait, but maybe I made a mistake in the direction. Let's check again.

Wait, if the base is longer, the included angle is larger. So, base 7 is longer than base 6, so the included angle for base 7 (31°) should be larger than the included angle for base 6 ((x - 3)°). So, (x - 3) < 31, so x < 34. And the angle (x - 3) must be positive, so x - 3 > 0 => x > 3. But also, the angle (x - 3) must be less than 180, but since it's part of a triangle, it's already less than 180, but in this case, since the other angle is 31°, the sum (x - 3) + 31 must be less than 180, so (x - 3) < 149, so x < 152, but that's a much larger upper bound. But the Hinge Theorem gives a tighter bound. Wait, but maybe the problem is that the two angles are part of a single triangle? Wait, no, the diagram shows two triangles with a common side (the altitude), and the two sides with tick marks are congruent. So, the two triangles are separate, but adjacent. So, the angle (x - 3)° and 31° are angles in their respective triangles.

Wait, but maybe the correct application is that the side opposite the larger angle is longer. So, in a triangle, the larger angle is opposite the longer side. So, in the triangle with sides: congruent side (a), altitude (h), and base 6: the angle opposite base 6 is (x - 3)°? No, the angle at the top is (x - 3)°, so the side opposite that angle is the base, 6. In the other triangle, the angle at the top is 31°, and the side opposite is 7. So, in a triangle, the larger side is opposite the larger angle. So, since 7 > 6, the angle opposite 7 (31°) should be larger than the angle opposite 6 ((x - 3)°). Therefore, 31° > (x - 3)°, so x - 3 < 31 => x < 34. And the angle (x - 3)° must be greater than 0°, so x - 3 > 0 => x > 3. Therefore, the range is 3 < x < 34.

Wait, but let's verify with the Law of Cosines. Suppose x = 4, then (x - 3) = 1°, so cos(1°) ≈ 0.9998, and cos(31°) ≈ 0.8572. Then, 2ah (0.9998 - 0.8572) = 2ah (0.1426) = 13 => 2ah = 13 / 0.1426 ≈ 91.16, which is positive, so that works. If x = 33, then (x - 3) = 30°, cos(30°) ≈ 0.8660, which is greater than cos(31°) ≈ 0.8572, so 2ah (0.8660 - 0.8572) = 2ah (0.0088) = 13 => 2ah = 13 / 0.0088 ≈ 1477.27, which is positive, so that works. If x = 3, then (x - 3) = 0°, cos(0°) = 1, so 2ah (1 - 0.8572) = 2ah (0.1428) = 13 => 2ah ≈ 91.04, but the angle can't be 0°, so x > 3. If x = 34, then (x - 3) = 31°, so cos(31°) = cos(31°), so 2ah (cos(31°) - cos(31°)) = 0 = 13, which is impossible, so x < 34. Therefore, the range is 3 < x < 34.

Step2: Verify the bounds

- Lower bound: \(x > 3\) because the angle \((x - 3)^\circ\) must be positive (an angle in a triangle cannot be zero or negative).
- Upper bound: \(x < 34\) because by the Hinge Theorem (or Law of Cosines), the included angle opposite the shorter base (6) must be smaller than the included angle opposite the longer base (7), so \((x - 3) < 31\), leading to \(x < 34\).

Answer:

The range is \(3 < x < 34\). So the first box is \(3\) and the second box is \(34\).