## Intermediate Algebra (12th Edition)

$(10x-20)^o=130^o ,\\ (x+15)^o=30^o ,\\ \left( x+5 \right)^o=20^o$
$\bf{\text{Solution Outline:}}$ Equate the sum of the angles to $180$ and solve for $x.$ Then substitute the value of $x$ in the following given angles: \begin{array}{l}\require{cancel} (10x-20)^o ,\\ (x+15)^o ,\\ \left( x+5 \right)^o .\end{array} $\bf{\text{Solution Details:}}$ Since the sum of the angles of a triangle is always $180^o,$ then equate the sum of the given angles to $180.$ That is, \begin{array}{l}\require{cancel} (10x-20)+(x+15)+\left( x+5 \right)^o =180 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 10x+x+x=180+20-15-5 \\\\ 12x=180 \\\\ x=\dfrac{180}{12} \\\\ x=15 .\end{array} Substituting $x= 15$ in the angle $(10x-20)^o$ results to \begin{array}{l}\require{cancel} (10\cdot15-20)^o \\\\= (150-20)^o \\\\= 130^o .\end{array} Substituting $x= 15$ in the angle $(x+15)^o$ results to \begin{array}{l}\require{cancel} (15+15)^o \\\\= 30^o .\end{array} Substituting $x= 15$ in the angle $(x+5)^o$ results to \begin{array}{l}\require{cancel} (15+5)^o \\\\= 20^o .\end{array} Hence, the measures of the angles of the triangle are \begin{array}{l}\require{cancel} (10x-20)^o=130^o ,\\ (x+15)^o=30^o ,\\ \left( x+5 \right)^o=20^o .\end{array}