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Isosceles Triangle Angles

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by mattnyc15 » Wed Nov 04, 2015 4:04 pm
I'm confused by this questions:


Image


I understand DCB = 60, ACD= 20 and each triangle is an isosceles.

But wouldn't that the other angle is DCB 60?? and ACD 140?

I know X must be the angle on the other side too, but from here i'm confused. [/img]


OA: X= 10
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by theCEO » Wed Nov 04, 2015 5:00 pm
mattnyc15 wrote:I'm confused by this questions:


Image


I understand DCB = 60, ACD= 20 and each triangle is an isosceles.

But wouldn't that the other angle is DCB 60?? and ACD 140?

I know X must be the angle on the other side too, but from here i'm confused. [/img]
DC = DB ; DBC = DCB = 60
BDC = 180-(DBC+DCB) = 180 - 120 = 60

AD = DC ; ACD = CAD = 20
ADC = 180 - (ACD + CAD) = 180 - 40 = 140

ADB = 360 - (ADC + BDC) = 360 - 200 = 160

BAD = ABD
BAD = (180 - ADB)/2 = (180 - 160)/2 = 10

X = 10


OA: X= 10
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by mattnyc15 » Wed Nov 04, 2015 7:37 pm
Thanks! The 360 for the circle in the middle made it click.
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