Question

what are the measures of the exterior angles of the polygon shown? (3x + 10)° (4x - 10)° (2x + 5)° (x + 15)° 3x°

Explanation:

Step1: Use the exterior angle sum formula

The sum of the exterior angles of any polygon is \(360^{\circ}\). So, \((x + 15)+(4x-10)+(3x + 10)+(2x + 5)+3x=360\).

Step2: Combine like terms

Combine \(x\) terms: \(x+4x+3x+2x+3x = 13x\). Combine constant terms: \(15-10 + 10+5=20\). The equation becomes \(13x+20 = 360\).

Step3: Solve for \(x\)

Subtract \(20\) from both sides: \(13x=360 - 20=340\). Then \(x=\frac{340}{13}\approx26.15\).

Step4: Find each exterior - angle measure

• For \(\angle1=(x + 15)^{\circ}\), substitute \(x\): \(\angle1=\frac{340}{13}+15=\frac{340+195}{13}=\frac{535}{13}\approx41.15^{\circ}\).
• For \(\angle2=(4x-10)^{\circ}\), substitute \(x\): \(\angle2 = 4\times\frac{340}{13}-10=\frac{1360}{13}-10=\frac{1360 - 130}{13}=\frac{1230}{13}\approx94.62^{\circ}\).
• For \(\angle3=(3x + 10)^{\circ}\), substitute \(x\): \(\angle3=3\times\frac{340}{13}+10=\frac{1020}{13}+10=\frac{1020+130}{13}=\frac{1150}{13}\approx88.46^{\circ}\).
• For \(\angle4=(2x + 5)^{\circ}\), substitute \(x\): \(\angle4=2\times\frac{340}{13}+5=\frac{680}{13}+5=\frac{680 + 65}{13}=\frac{745}{13}\approx57.31^{\circ}\).
• For \(\angle5 = 3x^{\circ}\), substitute \(x\): \(\angle5=3\times\frac{340}{13}=\frac{1020}{13}\approx78.46^{\circ}\).

Answer:

\(\angle1\approx41.15^{\circ},\angle2\approx94.62^{\circ},\angle3\approx88.46^{\circ},\angle4\approx57.31^{\circ},\angle5\approx78.46^{\circ}\)