Question

given: - $overleftrightarrow{tv}$ is parallel to $overleftrightarrow{wx}$. - the distance between $overleftrightarrow{tv}$ and $overleftrightarrow{wx}$ is always 6 units. - $overline{vw}$ is 10 units long. - $mangle 1 = 36.87^circ$ u was asked to find the measure of $angle 2$ and explain his reasoning. image of geometric figure with points t, u, v, w, x, y fill in the blanks in u’s solution. - we were given the distance between the parallel lines, so both $overline{uw}$ and $overline{yv}$ have a length of $\boxed{6}$ units. - $\triangle vuw$ and $\triangle wyv$ share a side, and we were given that this side is 10 units long. - both $\triangle vuw$ and $\triangle wyv$ are right triangles, so we can use the pythagorean theorem to find the lengths of $overline{vu}$ and $overline{wy}$. - $overline{vu}$ and $overline{wy}$ both have a length of $\boxed{10}$ units. - $\triangle vuw cong \triangle wyv$ by the side - side - side congruency postulate. - $mangle 2=square^circ$ since $angle 1$ and $angle 2$ are corresponding parts of congruent triangles.

Explanation:

Step1: Recall Congruent Triangles Property

Since \(\triangle VUW \cong \triangle WYV\) (by SSS congruence), their corresponding angles are equal.

Step2: Identify Corresponding Angles

\(\angle 1\) and \(\angle 2\) are corresponding parts of congruent triangles. Given \(m\angle 1 = 36.87^\circ\), so \(m\angle 2\) should be equal to \(m\angle 1\).

Answer:

\(36.87\)