MATH 03, WORKSHEET 6 KEY Problem 1. Given a

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MATH 03, WORKSHEET 6 KEY Problem 1. Given a Powered By Docstoc
					                               MATH 03, WORKSHEET 6 KEY

           WORKSHEET BY JESSICA ABARCA, SOLUTIONS BY EDOLFO GARZA-LICUDINE



Problem 1. Given √ right triangle with C = 90◦ , find...
                 a
First find CB = 2 2. Then:
          √                                  √
sin(A) = 2 3 3    cos(A) = 13      tan(A) = 2 2
            √                                        √
           3 2                                        2
csc(A) =    4       sec(A) = 3           cot(A) =    4


Problem 2. Given a right triangle with C = 90◦ , find...
First find AC = 1 . Then:
               4
         √
           3
                                           √
sin(A) = 2       cos(A) = 12      tan(A) = 3
            √                                        √
           2 3                                        3
csc(A) =    3       sec(A) = 2           cot(A) =    3


   Problem 3. Carry out the indicated operations
a. 11 sin(A) cos(A)
b. −8 cos2 (A) sin(A)
c. The most obvious thing to do here is: sec4 (A) − 9
   For those of you feeling particularly clever, you may want to challenge yourself to obtain:
                          √             √
   (sec2 (A) + 3)(sec(A) + 3)(sec(A) − 3)

d. −1
Problem 4. Factor each expression:
a. (tan(B) + 9)(tan(B) − 1)
b. 3 sec(B) tan(B)(3 sec(B) + 2 tan(B))
Problem 5. Use the given information to determine the value of the remaining five trigonometric
functions
                                 √
a. The third side has a length of 7. Then:
                             √                 √
   sin(A) = 3
            4      cos(A) = 47       tan(A) = 3 7 7
                                √                         √
             4                 4 7                         7
  csc(A) =   3      sec(A) =    7           cot(A) =      3
                                     √
b. The third side has a length of        22. Then:
             √                     √                          √
              22                      3                        66    22
  sin(B) =    5         cos(B) =     5       tan(B) =          3 =   3
              √                        √                       √
             5 22                    5 3                         66     3
  csc(B) =    22        sec(B) =       3       cot(B) =         22 =   22
                                  √
c. The third side has a length of √ 26. Then:
             √
               26
   sin(C) = 5 26       cos(C) = 26 26
                                          tan(C) = 5
            √
              26
                                √
   csc(C) = 5         sec(C) = 26         cot(C) = 1
                                                   5


  Date: May 13, 2009.
                                                          1
2                                                  E. GARZA-LICUDINE
                                          √
d. The third side has a length of              7. Then:
                √                        √                                 √
               2 7                        21           3                 2 3
    sin(D) =    7             cos(D) =    7    =       7      tan(D) =     3
               √                         √                              √
                 7                        21       7                      3
    csc(D) =    2             sec(D) =    3    =   3         cot(D) =   2


Problem 6. Write in terms of sine and cosine, then simplify.
a. sin(A) + cos(A)
     sin(A)    sin(A)
b. 1−cos(A) + 1+cos(A) = sin(A)[1+cos(A)]+sin(A)[1−cos(A)]
                                     1−cos2 (A)
c. cos 2 (B)

d. −(2 cos(A) + 1)
Problem 7. Given a right triangle, find:
                       √
a. BC = 30cm, AC = 3 3cm.
          √               √
b. AB = 323 3 cm, BC = 326 3 cm.
                                                                            √
Problem 8. Find height and base of the ladder triangle: Base = 9, height = 9 3.
Problem 9. The monument is measured to be about 555.5 feet high.
Problem 10. If the height of the building is h and the length of the antenna is l, we get:
                                                      h
                                           tan(A) =
                                                     30
                                                    l+h
                                          tan(B) =
                                                     30
So:
                                    l = 30 tan(B) − 30 tan(A)
                                                             600
Problem 11. Let x be the distance, then x =                tan(35◦ )

Problem 12. Find the area of the triangle:
               √
a. Area = 354 3
b. Area = 9/2
Problem 13. Distance ≈ 5400ft.
Problem 14. Evaluate:
                √
a. cos(315◦ ) = 22
                     √
b. cos(−315◦ ) =     2
                      2

c. sin(150◦ ) = 1
                2    √
d. sin(−225◦ ) =     2
                      2
                    √
e. cos(585◦ ) =    − 22
                       √
f. sin(−405◦ ) = −       2
                          2