Question

1. in \\( \triangle a b c \\), if \\( \overline{a c} \cong \overline{c b}, m \angle a=3 x+18, m \angle b=7 x-58 \\), and \\( m \angle c=2 x-8 \\), find \\( x \\) and the measure of each angle. \\( x= \\) \\( m \angle a= \\) \\( m \angle b= \\) \\( m \angle c= \\) 13. in \\( \triangle q r s \\), if \\( \overline{q r} \cong \overline{r s}, m \angle q=8 x-17, m \angle r=19 x+4 \\), and \\( m \angle s=5 x+1 \\), find \\( x \\) and the measure of each angle. \\( x= \\) \\( m \angle q= \\) \\( m \angle r= \\) \\( m \angle s= \\) 14. in \\( \triangle k l m \\), if \\( \angle k \cong \angle l, k l=9 x-40, l m=7 x-37 \\), and \\( k m=3 x+23 \\), find \\( x \\) and the measure of each side. \\( x= \\) \\( k l= \\) \\( l m= \\) \\( k m= \\)

Explanation:

Problem 12

Step1: Use the property of isosceles triangle

Since \(\overline{AC}\cong\overline{CB}\), then \(\angle A\cong\angle B\). So \(3x + 18=7x-58\).

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Step2: Calculate the measure of each angle

For \(\angle A\): \(m\angle A=3x + 18\), substitute \(x = 19\), \(m\angle A=3\times19+18=57 + 18=75^{\circ}\)
For \(\angle B\): \(m\angle B=7x-58\), substitute \(x = 19\), \(m\angle B=7\times19-58=133 - 58=75^{\circ}\)
For \(\angle C\): \(m\angle C=2x-8\), substitute \(x = 19\), \(m\angle C=2\times19-8=38 - 8=30^{\circ}\)

Step1: Use the property of isosceles triangle

Since \(\overline{QR}\cong\overline{RS}\), then \(\angle Q\cong\angle S\). So \(8x-17=5x + 1\)

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Step2: Calculate the measure of each angle

For \(\angle Q\): \(m\angle Q=8x-17\), substitute \(x = 6\), \(m\angle Q=8\times6-17=48-17 = 31^{\circ}\)
For \(\angle S\): \(m\angle S=5x + 1\), substitute \(x = 6\), \(m\angle S=5\times6+1=30 + 1=31^{\circ}\)
For \(\angle R\): \(m\angle R=19x + 4\), substitute \(x = 6\), \(m\angle R=19\times6+4=114+4=118^{\circ}\)

Step1: Use the property of isosceles triangle

Since \(\angle K\cong\angle L\), then \(LM=KM\). So \(7x-37=3x + 23\)

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Step2: Calculate the length of each side

For \(KL\): \(KL=9x-40\), substitute \(x = 15\), \(KL=9\times15-40=135-40 = 95\)
For \(LM\): \(LM=7x-37\), substitute \(x = 15\), \(LM=7\times15-37=105-37=68\)
For \(KM\): \(KM=3x + 23\), substitute \(x = 15\), \(KM=3\times15+23=45+23=68\)

Answer:

\(x = 19\), \(m\angle A=75^{\circ}\), \(m\angle B=75^{\circ}\), \(m\angle C=30^{\circ}\)

Problem 13