Question

list the side lengths of △abc in order from shortest to longest.

Explanation:

Step1: Sum of angles in a triangle

The sum of the interior angles of a triangle is \(180^\circ\). So we set up the equation: \(39p + (p + 67) + 74p = 180\).

Step2: Combine like terms

Combine the \(p\) terms: \(39p + p + 74p = 114p\), so the equation becomes \(114p + 67 = 180\).

Step3: Solve for \(p\)

Subtract 67 from both sides: \(114p = 180 - 67 = 113\)? Wait, that can't be right. Wait, maybe the angles are in degrees, but maybe I misread. Wait, no, the angles are \(39p^\circ\), \((p + 67)^\circ\), and \(74p^\circ\). Wait, let's check again. \(39p + p + 67 + 74p = 180\) → \(114p + 67 = 180\) → \(114p = 113\) → \(p=\frac{113}{114}\approx0.99\). That seems odd. Wait, maybe the angles are \(39^\circ\), \(p + 67^\circ\), and \(74^\circ\)? No, the diagram shows \(39p\), \(p + 67\), \(74p\). Wait, maybe a typo, but assuming the angle sum is 180, let's proceed.

Wait, maybe I made a mistake. Let's recalculate: \(39p + p + 74p = 114p\), \(114p + 67 = 180\) → \(114p = 113\) → \(p\approx0.99\). Then the angles are: \(39p\approx38.61^\circ\), \(p + 67\approx67.99^\circ\), \(74p\approx73.26^\circ\). Now, in a triangle, the larger the angle, the longer the opposite side. So angle at C: \(39p\approx38.61^\circ\) (opposite side AB), angle at B: \(p + 67\approx67.99^\circ\) (opposite side AC), angle at A: \(74p\approx73.26^\circ\) (opposite side BC). So the order of angles from smallest to largest: \(39p < p + 67 < 74p\) (wait, no, \(38.61 < 67.99 < 73.26\)), so the opposite sides: AB (opposite C) < AC (opposite B) < BC (opposite A). Wait, but maybe I messed up the angle labels. Let's check the triangle: vertex A, B, C. So side opposite A is BC, opposite B is AC, opposite C is AB. So angle at C: \(39p\), opposite AB; angle at B: \(p + 67\), opposite AC; angle at A: \(74p\), opposite BC. So if angles are \(39p < p + 67 < 74p\) (which they are with \(p\approx1\)), then sides: AB < AC < BC.

But wait, maybe the problem is that the angles are \(39^\circ\), \(p + 67^\circ\), \(74^\circ\), so sum is \(39 + 74 + p + 67 = 180\) → \(180 + p = 180\) → \(p = 0\), which is impossible. So maybe the original problem has angles \(39^\circ\), \(67^\circ\), and \(74^\circ\)? No, the diagram shows \(39p\), \(p + 67\), \(74p\). Maybe it's a different approach. Wait, maybe the user made a typo, but assuming the angle sum is 180, and solving for \(p\), then ordering the sides by opposite angles.

Alternatively, maybe the angles are \(39^\circ\), \(67^\circ\), and \(74^\circ\), sum is \(39 + 67 + 74 = 180\)! Oh! Wait, \(39 + 67 + 74 = 180\) (39+67=106, 106+74=180). So maybe the \(p\) is a typo, and the angles are \(39^\circ\), \(67^\circ\), \(74^\circ\). Then angle at C: \(39^\circ\) (opposite AB), angle at B: \(67^\circ\) (opposite AC), angle at A: \(74^\circ\) (opposite BC). So the sides opposite these angles: AB (opposite C: \(39^\circ\)), AC (opposite B: \(67^\circ\)), BC (opposite A: \(74^\circ\)). So the order from shortest to longest: AB, AC, BC.

Ah, that makes sense! So probably a typo, and the angles are \(39^\circ\), \(67^\circ\), \(74^\circ\) (since \(39 + 67 + 74 = 180\)). So then, in triangle ABC, side opposite angle C (39°) is AB, side opposite angle B (67°) is AC, side opposite angle A (74°) is BC. So the order of side lengths from shortest to longest is AB, AC, BC.

Answer:

AB, AC, BC (assuming angles are \(39^\circ\), \(67^\circ\), \(74^\circ\) due to angle sum being 180)