## Intermediate Algebra (12th Edition)

$(x-30)^o=60^0 ,\\ (2x-120)^o=60^o ,\\ \left( \dfrac{1}{2}x+15\right)^o=60^o$
$\bf{\text{Solution Outline:}}$ 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} (x-30)^o ,\\ (2x-120)^o ,\\ \left( \dfrac{1}{2}x+15\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} (x-30)+(2x-120)+\left( \dfrac{1}{2}x+15\right)=180 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} x+2x+\dfrac{1}{2}x=180+30+120-15 \\\\ 3x+\dfrac{1}{2}x=315 \\\\ 2\left( 3x+\dfrac{1}{2}x\right)=2(315) \\\\ 6x+x=630 \\\\ 7x=630 \\\\ x=\dfrac{630}{7} \\\\ x=90 .\end{array} Substituting $x= 90$ in the angle $(x-30)^o$ results to \begin{array}{l}\require{cancel} (90-30)^o \\\\= 60^o .\end{array} Substituting $x= 90$ in the angle $(2x-120)^o$ results to \begin{array}{l}\require{cancel} (2\cdot90-120)^o \\\\= (180-120)^o \\\\= 60^o .\end{array} Substituting $x= 90$ in the angle $\left(\dfrac{1}{2}x+15\right)^o$ results to \begin{array}{l}\require{cancel} \left(\dfrac{1}{2}\cdot90+15\right)^o \\\\= \left(45+15\right)^o \\\\= 60^o .\end{array} Hence, the measures of the angles of the triangle are \begin{array}{l}\require{cancel} (x-30)^o=60^0 ,\\ (2x-120)^o=60^o ,\\ \left( \dfrac{1}{2}x+15\right)^o=60^o .\end{array}