QUESTION IMAGE
Question
ws \triangle angle sum theorem\
find the measure of each angle indicated.
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Step1: Recall the triangle angle - sum theorem
The sum of the interior angles of a triangle is \(180^{\circ}\). For a triangle with angles \(A\), \(B\), and \(C\), \(A + B+C=180^{\circ}\). If we know two of the angles, say \(A\) and \(B\), then \(C = 180-(A + B)\)
Step2: Calculate the unknown angles
- For the first triangle with angles \(60^{\circ}\) and \(47^{\circ}\), the unknown angle \(y=180-(60 + 47)=180 - 107 = 73^{\circ}\)
- For the second triangle with angles \(60^{\circ}\) and \(58^{\circ}\), the unknown angle \(y=180-(60 + 58)=180 - 118 = 62^{\circ}\)
- For the third triangle (right - angled triangle with one angle \(90^{\circ}\) and another \(40^{\circ}\)), the unknown angle \(y=180-(90 + 40)=180 - 130 = 50^{\circ}\)
- For the fourth triangle with angles \(60^{\circ}\) and \(65^{\circ}\), the unknown angle \(y=180-(60 + 65)=180 - 125 = 55^{\circ}\)
- For the fifth triangle with angles \(85^{\circ}\) and \(35^{\circ}\), the unknown angle \(y=180-(85 + 35)=180 - 120 = 60^{\circ}\)
- For the sixth triangle with angles \(33^{\circ}\) and \(97^{\circ}\), the unknown angle \(y=180-(33 + 97)=180 - 130 = 50^{\circ}\)
- For the seventh triangle with angles \(30^{\circ}\) and \(70^{\circ}\), the unknown angle \(y=180-(30 + 70)=180 - 100 = 80^{\circ}\)
- For the eighth triangle with angles \(100^{\circ}\) and \(20^{\circ}\), the unknown angle \(y=180-(100 + 20)=180 - 120 = 60^{\circ}\)
- For the ninth triangle with angles \(34^{\circ}\) and \(90^{\circ}\), the unknown angle \(y=180-(34 + 90)=180 - 124 = 56^{\circ}\)
- For the tenth triangle with angles \(35^{\circ}\) and \(80^{\circ}\), the unknown angle \(y=180-(35 + 80)=180 - 115 = 65^{\circ}\)
- First, find the angle in the left - hand triangle: Let's call it \(x\). For the left - hand triangle with angles \(71^{\circ}\) and \(64^{\circ}\), \(x = 180-(71+64)=45^{\circ}\). Then, for the right - hand triangle with one angle \(110^{\circ}\) and \(x = 45^{\circ}\) (vertically opposite angles are equal), the unknown angle \(y=180-(110 + 45)=25^{\circ}\)
- First, find the angle in the left - hand triangle: Let's call it \(x\). For the left - hand triangle with angles \(81^{\circ}\) and \(20^{\circ}\), \(x = 180-(81 + 20)=79^{\circ}\). Then, for the right - hand triangle with one angle \(40^{\circ}\) and \(x = 79^{\circ}\) (linear pair: \(180 - 101=79\)), the unknown angle \(y=180-(40 + 79)=61^{\circ}\)
- First, find the angle in the left - hand triangle: Let's call it \(x\). For the left - hand triangle with angles \(58^{\circ}\) and \(62^{\circ}\), \(x = 180-(58 + 62)=60^{\circ}\). Then, for the right - hand triangle with angles \(47^{\circ}\) and \(60^{\circ}\), the unknown angle \(y=180-(47 + 60)=73^{\circ}\)
- First, find the non - labeled angle in the top triangle: Let's call it \(x\). For the top triangle with angles \(29^{\circ}\) and \(42^{\circ}\), \(x = 180-(29 + 42)=109^{\circ}\). The angle \(x\) and the angle adjacent to the unknown angle in the bottom triangle are vertically opposite, so they are equal. Then, for the bottom triangle with angles \(26^{\circ}\) and \(109^{\circ}\), the unknown angle \(y=180-(26 + 109)=45^{\circ}\)
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