Question

if $\angle 1\cong\angle 2\cong\angle 3$, $\angle 4\cong\angle 5$, if $\angle 1 = x$, and $\angle 4=x + 15^{circ}$, what is $m\angle 5$?
$m\angle 5?=$

Explanation:

Step1: Recall the sum of exterior angles of a polygon

The sum of the exterior angles of any polygon is \(360^{\circ}\). Here, \(\angle6 = 90^{\circ}\), \(\angle1=\angle2=\angle3 = x\), and \(\angle4=\angle5=x + 15^{\circ}\).

Step2: Set up the equation

We can write the equation \(x+x+x+(x + 15^{\circ})+(x + 15^{\circ})+90^{\circ}=360^{\circ}\).
Simplify the left - hand side: \(5x+120^{\circ}=360^{\circ}\).

Step3: Solve for \(x\)

Subtract \(120^{\circ}\) from both sides: \(5x=360^{\circ}-120^{\circ}=240^{\circ}\).
Divide both sides by \(5\): \(x = 48^{\circ}\).

Step4: Find \(m\angle5\)

Since \(\angle5=x + 15^{\circ}\), substitute \(x = 48^{\circ}\) into the formula. Then \(m\angle5=48^{\circ}+15^{\circ}\).

Answer:

\(63^{\circ}\)