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Question
question 10 of 10
what is the value of n?
a. 69°
b. 20°
c. 29°
d. 49°
Step1: Find the adjacent angle of \(160^{\circ}\)
The adjacent angle of \(160^{\circ}\) is \(180 - 160=20^{\circ}\) (linear - pair angles: sum to \(180^{\circ}\)).
Step2: Use the exterior - angle property
The exterior - angle property of a triangle states that the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
Let the exterior angle be \(131^{\circ}\), one non - adjacent interior angle be \(n^{\circ}\) and the other be \(20^{\circ}\). Then \(131=n + 20\).
Step3: Solve for \(n\)
Subtract \(20\) from both sides of the equation \(131=n + 20\).
\(n=131 - 20\)
\(n = 111\) (This is wrong. Let's use another property: The exterior angle of a polygon (here, we can consider the angles formed by the lines). The correct property is: If we consider the angles around the intersection of lines. Let's use the property of the exterior angle of a triangle - like figure formed by the angles.
Another way: The angle adjacent to \(131^{\circ}\) is \(180 - 131 = 49^{\circ}\).
Let's use the property that the sum of angles in a triangle - like (exterior - angle related) situation.
The angle adjacent to \(160^{\circ}\) is \(180 - 160=20^{\circ}\).
We know that \(n+20 = 49\) (using the property of the difference of angles formed by the intersecting lines)
\(n=49 - 20\)
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C. \(29^{\circ}\)