In the figure above, if $$AD$$ is parallel to $$BC,$$ then $$\angle ADC=\angle ADC=$$

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In the figure above, if $$AD$$ is parallel to $$BC,$$ then $$\angle ADC=\angle ADC=$$

by M7MBA » Sun Nov 28, 2021 1:16 am

00:00

A

B

C

D

E

Global Stats In the figure above, if $$AD$$ is parallel to $$BC,$$ then $$\angle ADC=\angle ADC=$$

A. $$11^{\circ}$$

B. $$22^{\circ}$$

C. $$33^{\circ}$$

D. $$46^{\circ}$$

E. $$134^{\circ}$$

Source: Princeton Review

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Re: In the figure above, if $$AD$$ is parallel to $$BC,$$ then $$\angle ADC=\angle ADC=$$

by [email protected] » Sun Nov 28, 2021 6:47 am
M7MBA wrote:
Sun Nov 28, 2021 1:16 am
1.jpg

In the figure above, if $$AD$$ is parallel to $$BC,$$ then $$\angle ADC=\angle ADC=$$

A. $$11^{\circ}$$

B. $$22^{\circ}$$

C. $$33^{\circ}$$

D. $$46^{\circ}$$

E. $$134^{\circ}$$

Source: Princeton Review
First, since angles in a triangle must add to 180°, we can see that the missing angle in the red triangle must be 180° - (x + 44)° Simplify this measurement to get (136 - x)° Finally, since AD is parallel to BC, we know that the two highlighted angles below must add to 180°. So, we can write: (136 - x)° + 2x° + 3x° = 180°
Simplify: 136 + 4x = 180
Subtract 136 from both sides: : 4x = 44
Solve: x = 11

Our goal is to find the measurement of ∠ADC
Since ∠ADC = 3x°, we can replace x with 11 to get: ∠ADC = 3x° = 3(11)° = 33°