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Area of Triangles
Please help quickly I'm desperate and running out of time.

 Jun 6, 2019
edited by AdamTaurus  Jun 6, 2019
 #1
avatar+106514 
+1

I think hectictar answered something like this a few days ago.....mine isn't as exquisite as hers

 

Area  =  (1/2)  (a) (b) sin C       (1)

 

And

 

   a            b                              a sin B

____  =  ____      so   b  =      ______           (2)

sinA       sin B                            sin A

 

Sub (2) into (1)     and we have that

 

Area  =  (1/2) (a) (a sin B) sin C             a^2 sin B sin C

             ___________________    =     ______________

                           ( sin  A)                             2  sin  A

 

You should be able to do the other two  using

 

Area =  (1/2) (b) (c) sin A           and   noting that          c  = b sin C /sin B

 

And

 

Area  = (1/2) (a) (c) sin B              and noting that   a  = c sin A / sin C

 

 

 

cool cool cool

 Jun 6, 2019

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