## Intermediate Algebra (12th Edition)

$(2x+7)^o=63^o ,\\ (x+61)^o=89^o ,\\ \left( x \right)^o=28^o$
$\bf{\text{Solution Outline:}}$ 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} (2x+7)^o ,\\ (x+61)^o ,\\ \left( x \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} (2x+7)+(x+61)+\left( x \right)=180 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 2x+x+x=180-7-61 \\\\ 4x=112 \\\\ x=\dfrac{112}{4} \\\\ x=28 .\end{array} Substituting $x= 28$ in the angle $(2x+7)^o$ results to \begin{array}{l}\require{cancel} (2\cdot28+7)^o \\\\= (56+7)^o \\\\= 63^o .\end{array} Substituting $x= 28$ in the angle $(x+61)^o$ results to \begin{array}{l}\require{cancel} (28+61)^o \\\\= 89^o .\end{array} Substituting $x= 28$ in the angle $(x)^o$ results to \begin{array}{l}\require{cancel} 28^o .\end{array} Hence, the measures of the angles of the triangle are \begin{array}{l}\require{cancel} (2x+7)^o=63^o ,\\ (x+61)^o=89^o ,\\ \left( x \right)^o=28^o .\end{array}