You will use the sine and cosine ratio to find the sides and angles of a right triangles Pardekooper

 sine  = opposite  Pardekooper hypotenuse cosine  = adjacent hypotenuse

 Lets try setting up for sin   C Pardekooper sin  = opposite hypotenuse sin  =

 Lets try setting up for cos   C Pardekooper cos  = adjacent hypotenuse cos  =

 Lets try setting up for cos   C Pardekooper cos  = adjacent hypotenuse cos  =

 Lets try setting up for sin   C Pardekooper sin  = opposite hypotenuse sin  =

  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 Remember this problem from o 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 sin  = opposite hypotenuse sin25 0 = o c c   c c sin25 0 = 15 c = 15 / sin

 Now lets find the hypotenuse  to the nearest whole number.  C Pardekooper cos  = adjacent hypotenuse cos   = c c cos58 0 = 7 7 cos58 0 c  c  13

Pardekooper Here comes the assignment

Assignment Workbook Page 403 all