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avatar+392 

Point $D$ is on side $AC$ of triangle $ABC$, $\angle ABD=15^{\circ}$ and $\angle DBC=50^{\circ}$. What is the measure of angle $BAD$, in degrees?

 Mar 26, 2020
 #1
avatar+111330 
+1

Note that angle BDC =  90 -  50  = 40°

 

And angle ADB is supplemental to  this =   180 - 40  =  140°

 

So  

 

BAD   + 15  + 140  =  180

 

BAD  +  155  = 180

 

BAD = 180 - 155  =  25°

 

 

cool cool cool

 Mar 26, 2020
 #2
avatar+724 
0

Angle  BAD = 180° - ( 15° + 50° +90° ) smiley

 Mar 26, 2020

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