Question

1. in \\( \triangle a b c, a c = 13 \mathrm { cm }, \angle a = 4 y - 71 ^ { \circ } \\), \\( \angle b = y + 7 ^ { \circ } \\), and \\( \angle c = 106 - 2 y ^ { \circ } \\). in \\( \triangle r s t \\), \\( r t = 2 x + 5 \mathrm { cm } \\) and \\( m \angle r = 113 ^ { \circ } \\). \\( \angle b \cong \angle s \\) and \\( \overline { a b } \cong \overline { r s } \\).

part a
what is \\( m \angle b \\) in degrees?
\\( m \angle b = \\)
part b
what is the value of \\( x \\)?
\\( x = \\)

Explanation:

Step1: Use the triangle angle - sum theorem

In \(\triangle ABC\), by the triangle angle - sum theorem \(\angle A+\angle B+\angle C = 180^{\circ}\).
Substitute \(\angle A=(4y - 71)^{\circ}\), \(\angle B=(y + 7)^{\circ}\), and \(\angle C=(106-2y)^{\circ}\) into the equation:
\((4y - 71)+(y + 7)+(106-2y)=180\)

Step2: Simplify the left - hand side of the equation

Combine like terms: \((4y+y - 2y)+(-71 + 7+106)=180\)
\(3y+42 = 180\)

Step3: Solve for \(y\)

Subtract \(42\) from both sides: \(3y=180 - 42\), so \(3y=138\)
Divide both sides by \(3\): \(y=\frac{138}{3}=46\)

Step4: Find \(m\angle B\) (for Part A)

Since \(\angle B=(y + 7)^{\circ}\) and \(y = 46\), then \(m\angle B=(46 + 7)^{\circ}=53^{\circ}\)

Step5: Prove \(\triangle ABC\cong\triangle RST\) (for Part B)

Since \(\angle B\cong\angle S\), \(\overline{AB}\cong\overline{RS}\), and \(\angle A=(4\times46-71)^{\circ}=113^{\circ}=\angle R\) (by ASA congruence criterion \(\triangle ABC\cong\triangle RST\))
Then \(AC = RT\) (corresponding parts of congruent triangles are equal)
Given \(AC = 13\) and \(RT=(2x + 5)\)
Set up the equation \(2x+5=13\)

Step6: Solve for \(x\) (for Part B)

Subtract \(5\) from both sides: \(2x=13 - 5\), so \(2x=8\)
Divide both sides by \(2\): \(x = 4\)

Answer:

Part A: \(m\angle B = 53^{\circ}\)
Part B: \(x = 4\)