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

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

4.  Lets try setting up for sin   C Pardekooper sin  = opposite hypotenuse sin  = 8 6 10 8

5.  Lets try setting up for cos   C Pardekooper cos  = adjacent hypotenuse cos  = 8 6 10 8

6.  Lets try setting up for cos   C Pardekooper cos  = adjacent hypotenuse cos  = 12 5 13 5

7.  Lets try setting up for sin   C Pardekooper sin  = opposite hypotenuse sin  = 12 5 13 5

8.   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 8.3 65 o 15

9.   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 = 15 65 o c c   c c sin25 0 = 15 c = 15 / sin25 0 15

10.  Now lets find the hypotenuse  to the nearest whole number.  C Pardekooper cos  = adjacent hypotenuse cos   = 58 0 7 7 c c cos58 0 = 7 7 cos58 0 c  c  13

11. Pardekooper Here comes the assignment

12. Assignment Workbook Page 403 all