Question

1. a triangle with one angle 80°, one side 52 + x, one side x + 62, vertex a.

Explanation:

Step1: Recall triangle angle sum

The sum of angles in a triangle is \(180^\circ\). So, \(80 + (x + 62) + (52 + x)=180\).

Step2: Simplify the equation

Combine like terms: \(80 + 62 + 52 + x + x = 180\) → \(194 + 2x = 180\).

Step3: Solve for x

Subtract 194: \(2x = 180 - 194 = -14\) → \(x = -7\). Wait, that seems odd. Wait, maybe the sides? Wait, no, the labels: maybe the sides are \(52 + x\) and \(x + 62\), but angles? Wait, no, maybe the labels are angles? Wait, the diagram: the angle at the top left is \(80^\circ\), angle at A is \(x + 62\), and the other angle is... Wait, maybe I misread. Wait, the sides? No, the notation: \(52 + x\) and \(x + 62\) as angles? Wait, no, maybe the triangle has angles \(80^\circ\), \(x + 62\) degrees, and the third angle is equal to \(52 + x\) degrees? Wait, no, in a triangle, angles sum to \(180\). So:

\(80 + (x + 62) + (52 + x) = 180\)

\(80 + 62 + 52 + 2x = 180\)

\(194 + 2x = 180\)

\(2x = 180 - 194 = -14\)

\(x = -7\). That gives negative angle, which is impossible. So maybe the labels are sides, and it's an isosceles triangle? Wait, \(52 + x\) and \(x + 62\) are equal? Wait, \(52 + x = x + 62\) → \(52 = 62\), no. Wait, maybe the angles opposite those sides? Wait, no, the diagram: maybe the angle at the top is \(80^\circ\), and the two sides are \(52 + x\) and \(x + 62\), and angle A is \(x + 62\). Wait, maybe it's an isosceles triangle with two angles equal? Wait, if \(52 + x\) and \(x + 62\) are angles, but they can't be equal. Wait, maybe a typo, but proceed with the equation. Wait, maybe I made a mistake. Let's re-express:

\(80 + (x + 62) + (52 + x) = 180\)

\(2x + 194 = 180\)

\(2x = -14\)

\(x = -7\). Hmm. Maybe the problem is about sides, but the labels are angles. Wait, maybe the original problem is about a triangle with sides, but the notation is confusing. Wait, no, the user's diagram: the triangle has angle \(80^\circ\), side \(52 + x\), side \(x + 62\), and angle at A is \(x + 62\). Wait, maybe it's an isosceles triangle where two sides are equal, so \(52 + x = x + 62\), which is impossible. Wait, no, maybe the angles: if \(x + 62\) and \(52 + x\) are angles, but that's the same as \(x + 52\), so no. Wait, maybe the third angle is \(52 + x\), and angle at A is \(x + 62\), and the top angle is \(80\). So sum: \(80 + (x + 62) + (52 + x) = 180\). So that's correct. Then \(x = -7\). But angle can't be negative. Maybe the diagram has a typo, but mathematically, solving gives \(x = -7\).

Answer:

\(x = -7\)