# MATH 03, WORKSHEET 6 KEY Problem 1. Given a

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```					                               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

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