Question

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

Explanation:

Step1: Recall SAS Congruence

SAS (Side - Angle - Side) congruence criterion states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.

Step2: Analyze the first option (ST = 20 ft, UV = 14 ft, and \(m\angle UST=98^{\circ}\))

We know that \(TU\) is a common side? Wait, no, let's look at the triangles \(\triangle STU\) and \(\triangle VTU\). We see that \(TU\) is a common side. If \(ST = 20\) ft (which is equal to \(VT = 20\) ft as seen from the diagram), \(UV=14\) ft (equal to \(US = 14\) ft) and \(\angle UST=\angle UVT = 98^{\circ}\) (included angle between the two sides), this satisfies SAS.

Step3: Analyze the option \(m\angle UST = 98^{\circ}\) and \(m\angle TUV=45^{\circ}\)

Wait, no, let's check the option " \(m\angle UST = 98^{\circ}\) and \(m\angle TUV = 45^{\circ}\)" - no, wait the option " \(m\angle UST = 98^{\circ}\) and \(m\angle TUV = 45^{\circ}\)" is not correct. Wait, the option " \(UV = 14\) ft and \(m\angle TUV=45^{\circ}\)": \(UV = 14\) ft (so \(US=14\) ft), \(TU\) is common, and \(\angle TUV=\angle TUS = 45^{\circ}\) (included angle), so this satisfies SAS.

Wait, let's re - evaluate:

For \(\triangle STU\) and \(\triangle VTU\):

• Option 1: \(ST = 20\) ft, \(UV = 14\) ft, and \(m\angle UST=98^{\circ}\)
• \(ST = VT = 20\) ft (from diagram, \(VT = 20\) ft), \(US=UV = 14\) ft, and \(\angle UST=\angle UVT = 98^{\circ}\) (included angle between \(US - ST\) and \(UV - VT\)), so SAS holds.

• Option 2: \(UV = 14\) ft and \(m\angle TUV = 45^{\circ}\)
• \(US = UV=14\) ft, \(TU\) is common, and \(\angle TUS=\angle TUV = 45^{\circ}\) (included angle between \(US - TU\) and \(UV - TU\)), so SAS holds.

Wait, the original options:

1. \(m\angle STU = 37^{\circ}\) and \(m\angle VTU = 37^{\circ}\): This is AA, not SAS.

1. \(TU = 26\) ft: This is a single side, not helpful for SAS.

1. \(ST = 20\) ft, \(UV = 14\) ft, and \(m\angle UST=98^{\circ}\): \(ST = VT = 20\) ft, \(US = UV = 14\) ft, \(\angle UST=\angle UVT = 98^{\circ}\) (included angle), so SAS.

1. \(UV = 14\) ft and \(m\angle TUV = 45^{\circ}\): \(US = UV = 14\) ft, \(TU\) common, \(\angle TUS=\angle TUV = 45^{\circ}\) (included angle), so SAS.

1. \(m\angle UST = 98^{\circ}\) and \(m\angle TUV = 45^{\circ}\): Not SAS.

So the two correct options are:

• \(ST = 20\) ft, \(UV = 14\) ft, and \(m\angle UST=98^{\circ}\)

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

Answer:

• \(ST = 20\) ft, \(UV = 14\) ft, and \(m\angle UST=98^{\circ}\)
• \(UV = 14\) ft and \(m\angle TUV = 45^{\circ}\)