Question

find the measure of ∠1 in each figure.
13.
14.
15.
16.
17.
18.
find the measure of each angle.
19.
20.
find each unknown angle measure.
18.
19.
20.

Explanation:

Step1: Recall angle - sum properties

For a triangle, the sum of interior angles is 180°. For a polygon with \(n\) sides, the sum of interior angles is \((n - 2)\times180^{\circ}\), and the sum of exterior angles of any polygon is 360°.

Step2: Solve problem 14

The given polygon is a quadrilateral (\(n = 4\)), so the sum of interior angles is \((4 - 2)\times180^{\circ}=360^{\circ}\). Let the unknown interior angle adjacent to \(\angle1\) be \(x\). We have \(x+85^{\circ}+100^{\circ}+90^{\circ}=360^{\circ}\), so \(x = 360^{\circ}-(85^{\circ}+100^{\circ}+90^{\circ})=85^{\circ}\). Since \(\angle1\) and \(x\) are supplementary (linear - pair), \(m\angle1 = 180^{\circ}-85^{\circ}=95^{\circ}\).

Step3: Solve problem 17

The sum of the two non - adjacent interior angles of the triangle is equal to the exterior angle. The two non - adjacent interior angles of the triangle are \(90^{\circ}\) and \(70^{\circ}\). So \(m\angle1=90^{\circ}+70^{\circ}=160^{\circ}\).

Step4: Solve problem 19 (hexagon)

The sum of interior angles of a hexagon (\(n = 6\)) is \((6 - 2)\times180^{\circ}=720^{\circ}\). Then \(x+(x + 20^{\circ})+(x + 40^{\circ})+(2x-50^{\circ})+80^{\circ}+150^{\circ}=720^{\circ}\). Combine like terms: \(5x+240^{\circ}=720^{\circ}\). Subtract 240° from both sides: \(5x=480^{\circ}\), so \(x = 96^{\circ}\). Then \(2x-50^{\circ}=2\times96^{\circ}-50^{\circ}=142^{\circ}\), \(x + 40^{\circ}=96^{\circ}+40^{\circ}=136^{\circ}\), \(x + 20^{\circ}=96^{\circ}+20^{\circ}=116^{\circ}\).

Step5: Solve problem 20 (quadrilateral)

The sum of interior angles of a quadrilateral is 360°. So \(3x^{\circ}+6x^{\circ}+(2x + 20)^{\circ}+(x + 40)^{\circ}=360^{\circ}\). Combine like terms: \(12x+60^{\circ}=360^{\circ}\). Subtract 60° from both sides: \(12x=300^{\circ}\), so \(x = 25^{\circ}\). Then \(3x=3\times25^{\circ}=75^{\circ}\), \(6x=6\times25^{\circ}=150^{\circ}\), \(2x + 20^{\circ}=2\times25^{\circ}+20^{\circ}=70^{\circ}\), \(x + 40^{\circ}=25^{\circ}+40^{\circ}=65^{\circ}\).

Step6: Solve problem 20 (last figure)

The sum of interior angles of a pentagon (\(n = 5\)) is \((5 - 2)\times180^{\circ}=540^{\circ}\). So \(x^{\circ}+95^{\circ}+50^{\circ}+102^{\circ}+47^{\circ}=540^{\circ}\). Combine like terms: \(x+294^{\circ}=540^{\circ}\). Subtract 294° from both sides: \(x = 246^{\circ}\).

Answer:

1. \(m\angle1 = 95^{\circ}\)
2. \(m\angle1 = 160^{\circ}\)
3. \(x = 96^{\circ}\), \(2x-50^{\circ}=142^{\circ}\), \(x + 40^{\circ}=136^{\circ}\), \(x + 20^{\circ}=116^{\circ}\)
4. \(x = 25^{\circ}\), \(3x=75^{\circ}\), \(6x=150^{\circ}\), \(2x + 20^{\circ}=70^{\circ}\), \(x + 40^{\circ}=65^{\circ}\) (first quadrilateral); \(x = 246^{\circ}\) (last pentagon)