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in \\( \triangle q r s, m \angle q=(6 x+17)^{circ}, m \angle r=(x+5)^{circ} \\), and \\( m \angle s=(3 x-2)^{circ} \\). find \\( m \angle r \\).
m\\( \angle r= \\)
answer attempt 1 out of 2
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Explanation:

Step1: Use the quadrilateral angle - sum formula

The sum of the interior angles of a quadrilateral is \(360^{\circ}\). So, \(m\angle Q + m\angle R+m\angle S + m\angle QRS=360^{\circ}\). Assuming \(QRS\) is a quadrilateral (since the problem mentions \(\angle Q,\angle R,\angle S\) and we need to use the angle - sum property of a quadrilateral). If it is a parallelogram (a common type of quadrilateral in such problems, and if we assume \(QRS\) is a parallelogram, then \(m\angle Q + m\angle R= 180^{\circ}\) (adjacent angles of a parallelogram are supplementary). So, \((6x + 17)+(x + 5)=180\).

Step2: Solve the linear equation for \(x\)

Combine like terms: \(6x+x+17 + 5=180\), which simplifies to \(7x+22 = 180\).
Subtract 22 from both sides: \(7x=180 - 22=158\).
Divide both sides by 7: \(x=\frac{158}{7}\approx22.57\) (This approach is wrong. Let's assume it's a quadrilateral. The sum of interior angles of a quadrilateral \(A + B + C+D = 360^{\circ}\). If we assume it's a quadrilateral \(QRS\) (maybe a typo for a quadrilateral name, say \(PQRS\)) and no other information about parallel sides, we use \(m\angle Q+m\angle R+m\angle S+m\angle P=360^{\circ}\). But if we assume it's a triangle (typo in the problem, maybe \(\triangle QRS\) is wrong and it's a quadrilateral). Let's re - do.
Assume it's a quadrilateral. \(m\angle Q=(6x + 17)^{\circ}\), \(m\angle R=(x + 5)^{\circ}\), \(m\angle S=(3x-2)^{\circ}\), and let the fourth angle \(m\angle T\). If it's a quadrilateral, \(6x + 17+x + 5+3x-2+m\angle T=360\). But if we assume it's a parallelogram (a common case in such problems where two angles are given in terms of \(x\) and we can use the property of adjacent angles being supplementary). So \(m\angle Q+m\angle R = 180\) (adjacent angles of a parallelogram).
\((6x + 17)+(x + 5)=180\)
\(7x+22 = 180\)
\(7x=180 - 22\)
\(7x=158\) (Wrong again. Wait, another approach: sum of angles in a quadrilateral \(=360^{\circ}\). If we assume it's a quadrilateral and no other information, but if we assume two pairs of angles: if it's a parallelogram \(m\angle Q=m\angle S\) and \(m\angle R=m\angle T\) (opposite angles are equal) and \(m\angle Q+m\angle R = 180\).
Let's use the sum of angles of a quadrilateral formula: \(m\angle Q+m\angle R+m\angle S+m\angle T = 360\). If we assume \(m\angle T=m\angle Q\) and \(m\angle S=m\angle R\) (parallelogram property). Then \(2(6x + 17)+2(x + 5)=360\)
\(12x+34 + 2x+10=360\)
\(14x+44 = 360\)
\(14x=360 - 44\)
\(14x=316\)
\(x = 22.57\) (still not an integer). Let's check the problem again. Maybe it's a triangle (typo). If it's a triangle \(m\angle Q+m\angle R+m\angle S=180\)
\((6x + 17)+(x + 5)+(3x-2)=180\)

Step2: Solve the equation

Combine like terms: \((6x+x + 3x)+(17 + 5-2)=180\)
\(10x+20 = 180\)
Subtract 20 from both sides: \(10x=180 - 20=160\)
Divide both sides by 10: \(x = 16\)

Step3: Find \(m\angle R\)

Substitute \(x = 16\) into \(m\angle R=(x + 5)^{\circ}\)
\(m\angle R=(16 + 5)^{\circ}=21^{\circ}\)

Answer:

\(21\)