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in △vwx, (overline{vw} cong overline{xv}) and (mangle x = 22^circ). find (mangle v).
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Explanation:

Step1: Recall the property of isosceles triangles

In a triangle, if two sides are congruent ($\overline{WV}\cong\overline{XV}$), then the triangle is isosceles. In an isosceles triangle $\triangle VWX$, the base - angles opposite the equal sides are equal. But here, we use the angle - sum property of a triangle. The sum of the interior angles of a triangle is $180^{\circ}$. Let $m\angle V = m\angle W=x$ (since $\overline{WV}\cong\overline{XV}$, the angles opposite these sides are equal). We know that $m\angle X = 22^{\circ}$.

Step2: Apply the angle - sum formula

By the angle - sum property of a triangle $m\angle V+m\angle W + m\angle X=180^{\circ}$. Substituting $m\angle V = m\angle W=x$ and $m\angle X = 22^{\circ}$, we get $x + x+22^{\circ}=180^{\circ}$.

Step3: Solve the equation

Combine like terms: $2x+22^{\circ}=180^{\circ}$. Subtract $22^{\circ}$ from both sides: $2x=180^{\circ}- 22^{\circ}=158^{\circ}$. Divide both sides by 2: $x=\frac{158^{\circ}}{2}=79^{\circ}$.

Answer:

$79^{\circ}$