Question

angular problems
solve these problems.

1. refer to figure 21 - 36 and determine the values of angles 1 through 10.

Explanation:

Step1: Find \(\angle1\)

Use the property that the sum of angles around a point is \(360^{\circ}\). For the angles \(83^{\circ}\), \(\angle1\), \(\angle2\) and \(132^{\circ}\) around a point. Also, \(\angle1\) and the \(132^{\circ}\) angle are supplementary (linear - pair). \(\angle1=180^{\circ}- 132^{\circ}=48^{\circ}\)

Step2: Find \(\angle2\)

\(\angle2\) and \(83^{\circ}\) are supplementary (linear - pair). \(\angle2 = 180^{\circ}-83^{\circ}=97^{\circ}\)

Step3: Find \(\angle3\)

\(\angle3\) and \(83^{\circ}\) are vertical angles. So, \(\angle3 = 83^{\circ}\)

Step4: Find \(\angle4\)

\(\angle4\) and \(132^{\circ}\) are vertical angles. So, \(\angle4=132^{\circ}\)

Step5: Find \(\angle5\)

First, find the angle adjacent to \(145^{\circ}\) (let's call it \(x\)), \(x = 180^{\circ}-145^{\circ}=35^{\circ}\). Then, for the triangle - like formation with angles \(98^{\circ}\), \(x\) and \(\angle5\) (using the sum of angles in a triangle \(180^{\circ}\)), \(\angle5=180^{\circ}-98^{\circ}-35^{\circ}=47^{\circ}\)

Step6: Find \(\angle6\)

\(\angle6\) and \(145^{\circ}\) are supplementary (linear - pair). \(\angle6 = 180^{\circ}-145^{\circ}=35^{\circ}\)

Step7: Find \(\angle7\)

\(\angle7\) and \(98^{\circ}\) are supplementary (linear - pair). \(\angle7=180^{\circ}-98^{\circ}=82^{\circ}\)

Step8: Find \(\angle8\)

\(\angle8\) and \(\angle5\) are vertical angles. So, \(\angle8 = 47^{\circ}\)

Step9: Find \(\angle9\)

\(\angle9\) and \(117^{\circ}\) are supplementary (linear - pair). \(\angle9=180^{\circ}-117^{\circ}=63^{\circ}\)

Step10: Find \(\angle10\)

\(\angle10\) and \(117^{\circ}\) are vertical angles. So, \(\angle10 = 117^{\circ}\)

Answer:

\(\angle1 = 48^{\circ}\), \(\angle2=97^{\circ}\), \(\angle3 = 83^{\circ}\), \(\angle4=132^{\circ}\), \(\angle5 = 47^{\circ}\), \(\angle6=35^{\circ}\), \(\angle7 = 82^{\circ}\), \(\angle8=47^{\circ}\), \(\angle9 = 63^{\circ}\), \(\angle10=117^{\circ}\)