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Question 1 of 4
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The sum of the interior angles in a quadrilateral is 360 ° 360 °
Since the interior angles of a quadrilateral add to 360 ° , 360 ° , 360 ° . 360 ° . x x
30 x - 2 + 62 + 62 + 118 30 x − 2 + 62 + 62 + 118 = = 360 360
30 x + 240 30 x + 240 = = 360 360 Simplify
30 x + 240 30 x + 240 - 240 − 240 = = 360 360 - 240 − 240 Subtract 240 240
30 x 30 x = = 120 120
30 x 30 x ÷ 30 ÷ 30 = = 120 120 ÷ 30 ÷ 30 Divide both sides by 30 30
x x = = 4
Question 2 of 4
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Supplementary angles are when two angles have a sum of 180 ° .
Opposite angles of a parallelogram have equal values.
Find the missing supplementary angle, which is equal to the value of x
First, we can see from the diagram that the exterior angle 75 ° ∠ A D C
Since supplementary angles add to 180 ° , 180 ° . a
∠ A D C + 75 = 180
∠ A D C + 75 - 75 = 180 - 75 Subtract 75
∠ A D C = 105 °
Finally, the angle ∠ A D C x
Since opposite angles on a parallelogram are equal, ∠ x = 105 °
Question 3 of 4
Find the value of a b c
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Alternate Angles are equal.
The sum of the interior angles in a triangle is 180 °
Opposite angles of a parallelogram have equal values.
First, we can see from the diagram that 35 ° b
Next, since the interior angles of a triangle add to 180 ° , 180 ° . c
b + c + 113 = 180
35 + c + 113 = 180 Plug in the known values
c + 148 = 180 Simplify
c + 148 - 148 = 180 - 148 Subtract 148
∠ c = 32 °
Finally, the angle 113 ° a
Since opposite angles on a parallelogram are equal, ∠ a = 113 °
Question 4 of 4
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An Isosceles Triangle has two congruent sides (the two sides with dashes) and the two base angles are equal.
Supplementary angles are when two angles have a sum of 180 ° .
The sum of the interior angles in a quadrilateral is 360 °
To solve for ∠ B E D ∠ E B C 360 °
First, since the base angles in an isosceles triangle are equal, ∠ A B E 70 °
Next, we can see from the diagram that the angle ∠ A B E ∠ E B C
Since supplementary angles add to 180 ° , 180 ° . ∠ E B C
∠ E B C + ∠ A B E = 180
∠ E B C + 70 = 180 Plug in the known values
∠ E B C + 70 - 70 = 180 - 70 Subtract 70
∠ E B C = 110 °
Finally, since the interior angles of a quadrilateral add to 360 ° , 360 ° . ∠ B E D
∠ B E D + ∠ E B C + 109 + 84 = 360
∠ B E D + 110 + 109 + 84 = 360 Plug in the known lengths
∠ B E D + 303 = 360 Simplify
∠ B E D + 303 - 303 = 360 - 303 Subtract 303
∠ B E D = 57 °
