Question

given: a // b, m∠1=(x + 3y)°, m∠2=(2x + 30)°, m∠3=(5y + 20)°. what is m∠1? a 40° b 70° c 10° d 110°

Explanation:

Step1: Use corresponding - angles property

Since \(a\parallel b\), \(\angle2\) and \(\angle3\) are corresponding angles, so \(m\angle2 = m\angle3\). Then we have the equation \(2x + 30=5y + 20\). Rearranging gives \(2x-5y=- 10\).

Step2: Use linear - pair property

\(\angle1\) and \(\angle2\) form a linear - pair, so \(m\angle1 + m\angle2=180^{\circ}\), that is \((x + 3y)+(2x + 30)=180\). Simplify to get \(3x+3y = 150\), and further to \(x + y=50\), so \(x = 50 - y\).

Step3: Substitute \(x\) into the first equation

Substitute \(x = 50 - y\) into \(2x-5y=-10\). We get \(2(50 - y)-5y=-10\). Expand: \(100-2y-5y=-10\). Combine like - terms: \(100-(2y + 5y)=-10\), \(100 - 7y=-10\). Subtract 100 from both sides: \(-7y=-110\), then \(y = 10\).

Step4: Find \(x\)

Substitute \(y = 10\) into \(x = 50 - y\), we get \(x=40\).

Step5: Find \(m\angle1\)

Substitute \(x = 40\) and \(y = 10\) into \(m\angle1=(x + 3y)\). Then \(m\angle1=(40+3\times10)=70^{\circ}\).

Answer:

B. \(70^{\circ}\)