Question

the figure is not drawn to scale. $overline{ab}$ and $overline{cd}$ are straight lines: $angle b$ is three times as large as $angle a$ and $10^{circ}$ smaller than $angle c$. if $mangle bgd = 72^{circ}$, find $mangle dge$. $mangle dge=___^{circ}$ enter the answer

Explanation:

Step1: Set up equations based on angle - relationships

Let \(m\angle a=x\). Then \(m\angle b = 3x\) and \(m\angle c=3x + 10^{\circ}\). Since \(\angle BGD\) and \(\angle AGC\) are vertical angles, \(m\angle AGC=m\angle BGD = 72^{\circ}\). And \(m\angle AGC=m\angle a+m\angle b\), so \(x + 3x=72^{\circ}\).

Step2: Solve for \(x\)

Combining like - terms in the equation \(x + 3x=72^{\circ}\), we get \(4x=72^{\circ}\). Dividing both sides by 4, \(x=\frac{72^{\circ}}{4}=18^{\circ}\).

Step3: Find \(m\angle DGE\)

\(\angle DGE\) and \(\angle AGC\) are vertical angles. We know \(m\angle a=x = 18^{\circ}\), \(m\angle b = 3x=3\times18^{\circ}=54^{\circ}\), and \(m\angle c=3x + 10^{\circ}=54^{\circ}+10^{\circ}=64^{\circ}\). Since \(\angle DGE\) and \(\angle AGC\) are vertical angles, and \(m\angle AGC=m\angle a+m\angle b\), \(m\angle DGE = 72^{\circ}\).

Answer:

\(72\)