Question

the frame of a bridge is constructed of triangles as shown. what additional information could you use to show that \\( \triangle stu \cong \triangle vtu \\) using sas? choose two correct answers. \\( \mathrm { st } = 20 \mathrm { ft }, \mathrm { uv } = 14 \mathrm { ft } \\), and \\( m \angle \mathrm { ust } = 98 ^ { \circ } \\) \\( \mathrm { tu } = 26 \mathrm { ft } \\) \\( m \angle \mathrm { stu } = 37 ^ { \circ } \\) and \\( m \angle \mathrm { vtu } = 37 ^ { \circ } \\) \\( m \angle \mathrm { ust } = 98 ^ { \circ } \\) and \\( m \angle \mathrm { tuv } = 45 ^ { \circ } \\) \\( \mathrm { uv } = 14 \mathrm { ft } \\) and \\( m \angle \mathrm { tuv } = 45 ^ { \circ } \\)

Explanation:

Step1: Recall the SAS (Side - Angle - Side) congruence criterion

For two triangles \(\triangle STU\) and \(\triangle VTU\), we already have \(TU = TU\) (common side).

Step2: Analyze each option

• Option 1: \(ST = 20\mathrm{ft}, UV = 14\mathrm{ft},m\angle UST=98^{\circ}\)

This does not give a side - angle - side relationship for \(\triangle STU\) and \(\triangle VTU\) as the sides and angles are not in the correct correspondence for SAS.

• Option 2: \(TU = 26\mathrm{ft}\)

Just knowing the length of \(TU\) (which is common) does not provide new information for SAS.

• Option 3: \(m\angle STU = 37^{\circ}\) and \(m\angle VTU=37^{\circ}\)

We have \(ST\) (if we consider the sides), \( \angle STU=\angle VTU\) (the angle) and \(TU = TU\) (the common side). If we assume \(ST = UV\) (from the figure, \(ST\) and \(UV\) are in a position where if we use the angle between \(ST - TU\) and \(UV - TU\)). But more precisely, if we consider the two triangles \(\triangle STU\) and \(\triangle VTU\), \(TU\) is common. If \(\angle STU=\angle VTU\), and if we can get the adjacent sides. From the figure, \(SU = 14\mathrm{ft}\), if \(UV = 14\mathrm{ft}\) (by the problem's structure, assume \(SU\) and \(UV\) are corresponding sides). Then \(SU = UV\), \(\angle STU=\angle VTU\), \(TU = TU\) (SAS).

• Option 4: \(m\angle UST = 98^{\circ}\) and \(m\angle TUV = 45^{\circ}\)

This does not give a side - angle - side relationship for \(\triangle STU\) and \(\triangle VTU\) as the angles are not the included angles between the corresponding sides.

• Option 5: \(UV = 14\mathrm{ft}\) and \(m\angle TUV = 45^{\circ}\)

We have \(SU = 14\mathrm{ft}\) (from the figure), \(UV = 14\mathrm{ft}\), \( \angle TUV\) (if we consider the angle between \(UV - TU\)) and \(TU = TU\) (common side). If we assume the angle between \(SU - TU\) and \(UV - TU\) (by the problem's structure, \(\angle TUS=\angle TUV\) (vertically opposite or by construction in the bridge frame). Then \(SU = UV\), \(\angle TUS=\angle TUV\), \(TU = TU\) (SAS)

Answer:

\(m\angle STU = 37^{\circ}\) and \(m\angle VTU = 37^{\circ}\); \(UV = 14\mathrm{ft}\) and \(m\angle TUV = 45^{\circ}\)