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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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MATH 03, WORKSHEET 6 KEY Problem 1. Given a, Fall 2008, UC Santa Cruz, PDF archives, esercizi di matematica, level physics, search engine, Math Worksheet, Algebra Worksheet

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posted: | 5/30/2009 |

language: | English |

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