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In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 45^\circ,$ $\angle ABC = 45^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$

 Mar 12, 2024
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45  A

         24

 

   C           D              B 45

 

Angle CAD  =22.5   Angle CDA = 67.5   Angle ACD =  90

AC / sin (CDA)  = AD / sin (ACD)

AC/ sin (67.5) = 24 / 1

AC = 24sin (67.5)

 

BAC = 45    = ABC

So  BC  = AC

 

Area =  (1/2)(BC * AC)  =  (1/2)[ 24 sin (67.5)] ^2  ≈   245.8 

 

 

cool cool cool

 Mar 12, 2024

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