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 How many parts are there to a triangle ? a b c  C Pardekooper AA BB CC

 tan  = opposite  Pardekooper adjacent

 Lets try setting up for tan   C Pardekooper tan  = opposite adjacent tan  =

 Lets try setting up for tan   C Pardekooper tan  = opposite adjacent tan  =

 Lets try setting up for tan   C Pardekooper tan  = opposite adjacent tan  =

 Lets try setting up for tan   C Pardekooper tan  = opposite adjacent tan  =

Pardekooper   Solve right triangle ABC if b=32,  A=25 o, and  C=90 o a b=32 c = 25 o = 90 o The sum of the angles of a triangle is 180 o. = 65 o A B C a c b AA BB CC  A+  B+  C = 180 o. 25 o +  B+90 o = 180 o.  B+115 o = 180 o  B = 65 o 65 o 32

  Solve right triangle ABC if b=32,  A=25 o, and  C=90 o a b=32 c = 25 o = 90 o = 65 o A B C a c =15 32 AA BB CC Pardekooper tan  = opposite adjacent tan25 0 = a 65 o   32 32tan25 0 = a 15 = a 15

  Solve right triangle ABC if b=32,  A=25 o, and  C=90 o a b=32 c = 25 o = 90 o = 65 o A B C c =15 32 AA BB CC Pardekooper We will find c tomorrow 65 o 15

 Now lets find  A  to the nearest degree.  C Pardekooper tan  = opposite adjacent tan  =  = tan  = 58 0

 Now lets find  B  to the nearest degree.  C Pardekooper tan  = opposite adjacent tan  =  = tan  = 32 0

Pardekooper Here comes the assignment

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