Question

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Explanation:

Step1: Identify Vertical Angles

Vertical angles are equal, so \(3x + 30 = x + 115\).

Step2: Solve for \(x\)

Subtract \(x\) from both sides: \(2x + 30 = 115\).
Subtract 30: \(2x = 85\).
Divide by 2: \(x = \frac{85}{2} = 42.5\). Wait, maybe I misread the diagram. Wait, maybe the lines are parallel? Wait, no, maybe the triangle has angles. Wait, no, the vertical angle: Wait, maybe the angle at D: \(3x + 30\) and the angle in the triangle? Wait, no, maybe the exterior angle. Wait, let's re - examine. If we consider the triangle \(DEF\), and the angle at D is \(3x + 30\), and the other angles are \(2x - 35\) and \(x + 115\). Wait, no, the sum of angles in a triangle is \(180^{\circ}\). Wait, maybe the angle \(3x + 30\) is equal to the sum of the two remote interior angles (exterior angle theorem). So \(3x+30=(2x - 35)+(x + 115)\).

Step3: Apply Exterior Angle Theorem

The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So \(3x + 30=(2x-35)+(x + 115)\).
Simplify the right - hand side: \(2x-35+x + 115=3x + 80\).
So we have the equation \(3x+30 = 3x + 80\).
Subtract \(3x\) from both sides: \(30=80\), which is a contradiction. Wait, maybe the vertical angle is equal to \(2x - 35\) or \(x + 115\). Wait, maybe the angle \(x\) and \(3x + 30\) are supplementary? No, maybe I made a mistake in identifying the angles. Wait, let's start over.
Suppose the two lines are such that the angle \(3x + 30\) and the angle in the triangle (let's say the angle at F is \(2x-35\) and at E is \(x + 115\)) and the angle at D ( \(3x + 30\)) is equal to the sum of the angles at E and F (exterior angle theorem). So \(3x+30=(x + 115)+(2x-35)\).
Simplify the right - hand side: \(x+115 + 2x-35=3x + 80\).
So \(3x+30 = 3x + 80\), which gives \(30 = 80\), impossible. Wait, maybe the angle \(3x + 30\) is equal to \(x + 115\) (vertical angles or alternate angles). Let's try that. \(3x+30=x + 115\).
Subtract \(x\) from both sides: \(2x+30 = 115\).
Subtract 30 from both sides: \(2x=85\), \(x = 42.5\). But then check the other angle \(2x-35=2\times42.5-35 = 85 - 35=50\), and \(x + 115=42.5+115 = 157.5\), and \(3x + 30=3\times42.5+30 = 127.5+30 = 157.5\), which matches. Wait, maybe the angle \(3x + 30\) and \(x + 115\) are equal (vertical angles or corresponding angles). Let's solve \(3x+30=x + 115\).
\(3x - x=115 - 30\)
\(2x = 85\)
\(x=\frac{85}{2}=42.5\)
Wait, but let's check the sum of angles in the triangle. The angles in the triangle would be \(x + 115=42.5+115 = 157.5\), \(2x-35=2\times42.5-35 = 50\), and the angle at D (if it's equal to \(x + 115\)) is \(157.5\). But \(157.5+50+157.5>180\), which is wrong. Wait, maybe the angle \(3x + 30\) is supplementary to \(x + 115\). \(3x+30+x + 115 = 180\)
\(4x+145 = 180\)
\(4x=35\)
\(x = 8.75\), which doesn't seem right.
Wait, maybe I misread the diagram. Let's assume that the angle \(3x + 30\) and \(2x-35\) are vertical angles. So \(3x+30=2x - 35\)
\(3x-2x=-35 - 30\)
\(x=-65\), which is not possible for an angle.
Alternatively, if \(3x + 30\) and \(x + 115\) are vertical angles: \(3x+30=x + 115\)
\(2x=85\)
\(x = 42.5\)
Then the angle \(2x-35=2\times42.5-35 = 50\), and the sum of angles in the triangle: \(x + 115+2x-35+ \text{angle at D}\). If angle at D is \(3x + 30 = 157.5\), then \(157.5+50+157.5 = 365\), which is wrong. Wait, maybe the diagram is such that the two lines are parallel, and we have alternate interior angles. Suppose the line with angle \(x\) and \(3x + 30\) is a transversal, and the triangle has angles \(2x-35\) and \(x + 115\). Maybe the angle \(x\) is equal to \(2x-35\) (alternate interior angles). Then \(x=2x - 35\), \(x = 35\). Then \(3x+30=3\times35 + 30=135\), and \(x + 115=35 + 115 = 150\), which doesn't match.
Wait, maybe the sum of angles in the triangle: \((2x-35)+(x + 115)+(3x + 30)=180\)
Combine like terms: \(2x+x+3x-35 + 115+30=6x + 110\)
Set equal to 180: \(6x+110 = 180\)
\(6x=70\)
\(x=\frac{70}{6}=\frac{35}{3}\approx11.67\), which also doesn't seem right.
Wait, maybe the angle \(3x + 30\) is an exterior angle, and the two interior angles are \(2x-35\) and \(x + 115\), but I made a mistake in the sign. Maybe \(3x+30=(x + 115)-(2x-35)\), but that would be \(3x+30=-x + 150\), \(4x=120\), \(x = 30\). Let's check: \(3x+30=120\), \(2x-35=25\), \(x + 115=145\), \(145 - 25 = 120\), which works (if we consider the exterior angle as the difference, but that's not the exterior angle theorem). The exterior angle theorem is sum, not difference. But if we assume that the angle \(x + 115\) is an exterior angle, then \(x + 115=(3x + 30)+(2x-35)\)
\(x + 115=5x - 5\)
\(120 = 4x\)
\(x = 30\)
Ah! This makes sense. Let's re - apply the exterior angle theorem correctly. If we consider angle \(x + 115\) as an exterior angle of the triangle with interior angles \(3x + 30\) and \(2x-35\), then by the exterior angle theorem, the exterior angle is equal to the sum of the two non - adjacent interior angles. So \(x + 115=(3x + 30)+(2x-35)\)
Simplify the right - hand side: \(3x+30+2x-35 = 5x-5\)
So we have the equation \(x + 115=5x-5\)
Subtract \(x\) from both sides: \(115 = 4x-5\)
Add 5 to both sides: \(120 = 4x\)
Divide both sides by 4: \(x = 30\)

Answer:

\(x = 30\)