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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Isosceles Triangle Angles
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I'm confused by this questions:
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
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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theCEO
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DC = DB ; DBC = DCB = 60mattnyc15 wrote:I'm confused by this questions:
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]
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
