Ichar Nicole Heffernan Grade 10 Trigonometry and Advanced Trigonometry Math Test Corrections Level 1/2 1. Find the unknown length in the following diagram. Give your answer to 2 decimal places: cos30 = x/15 x = 12.99 Initially, I had only gotten half of the two points for this question because I left my answer at 13. The reason that the right answer is 12.99 is because the question called for the answer to be rounded to two decimal places rather than be rounded to a whole number. 3. Find the unknown angle in the following diagram give your answers to 2 decimal places. sin0 = 7.3/12.4 0 = 36.065o Originally, my answer for this question was 19.79 degrees, however upon revision, I realized that the answer was actually 36.065 degrees. To be honest I’m not exactly sure where I got my original answer from but the reason that the right answer is 36.065 degrees is because we had the values of one of the angles and one of the sides therefore, we used the sin rule to find the remaining angle. 11. Find the area of the following triangle: A = 1/2 x 2.8 x 4.6 x sin37 A = 3.8756 km2 For this question, I had given the right numerical answer, however, I only received half of the point because I had not inserted km2 after the numerical value. It is important to always put the units at the end of your answer so one can know what you’re talking about. Level 3/4 12. From a point 120m horizontally from the base of a building, the angle of elevation to the top of the building is 34 degrees. Find the height of the building. x/sin34 = 120/sin56 Ichar Nicole Heffernan Grade 10 x = 80.95 m I had gotten this question wrong due to a silly mistake. In order o find the answer to this question you must isolate your x. In order to isolate the x you must put it over sin34 and make it equal to 120/sin56. Once you do this you must multiply both sides by sin 34 and this will end up equaling 80.95 m. 14. Find the size of the obtuse angle PQR by using trigonometry. 102 = 52 + 72 - (2 x 5 x 7 x cos0) 100 = 74 - 70cos0 26 = 70cos0 -0.371 = cos0 0 = 1.1907 Although I don’t think that this final answer is correct due to the fact that it is an extremely small number, I tried to complete this by using the cosine rule. 16. Solve the following on 0 < 0 < 2π a. sin 0 = 0.6224 = 0. 672 or 2.469 b. tan 0 = -5/3 = 120.397 c. cos 0 = -0.506 = 2.101 or 4.181 For this question I had tried to solve the question by using inverse sin, tan and cos. However, in order to find the answer to these questions you must use the unit circle and pi. For sin, you must first inverse it and then subtract your answer from pi and add it to pi. You must follow these same steps for cos. 17. Solve the following on 0 < 0 < 2π cos20 - 2cos0 + 1 = 0 (cos0 - 1)(cos0 - 1) = 0 cos0 - 1 = 0 cos0 = 1 cos = 0, 2π Ichar Nicole Heffernan Grade 10 For this question, I initially had not provided an answer. However, the reason why the answer above is correct is because I used factorization in order to fin the answer. I first factorized the left hand side and then simplified the question. 1 then moved to the right hand side, and I used inverse cos in order to find the final answer. Level 5/6 18. An observer at the top of a tower height 15 m sees a man due west of her at an angle of depression 31o. She sees another man due South at an angle of depression 17o. Find the distance between the men. Unfortunately I was not able to come to a conclusion for this question. 19. Bush walkers leave point P and walk in the direction 238o for 11.3 km to point Q. At Q they change direction to 107o and walk for 18.9 km to point R. How far is R from the starting point P? x2 = 11.32 + 18.12 - (2 x 11.3 x 18.1 x cos49) x2 = 332.34 x = 18.23 The reason that x equals 18.23 is due to the cosine rule. Through the cosine rule I was able to isolate the x and find the distance between P and R. Initially, I didn’t properly set up this question, however, upon further inspection I found that I could complete this question accurately. 21. Simplify: 4sin (π/2 - 0) - 2cos (- 0) 4cos0 - 2cos(-0) 4cos0 - 2(cos0) 4cos0 - 2cos0 = 2cos0 Ichar Nicole Heffernan Grade 10 Originally, I had not provided an answer for this question, however, after I looked back at the book, I realized that these were the rules that we learned in chapter 18. These rules included that cos(-0) = cos 0 and sin(-0) = -sin 0. 22. For each of the graphs below, establish a formula of the form y = A sin (Bx) + C y = 3sinx y = sinx - 2 Originally, I had not written understood what we were find the 11.3 km graph. an answer for this question on this test as I hadn’t supposed to do. However, now I understand that in order to answer to this question all one needs to do is read the P 23. Prove the x following trigonometric identity: 49 18.1 km (4sin0 + R 3cos0)2 + (3sin0 - 4cos0)2 = 25 If sin0 = x and cos 0 = y then : 16x2 + 12xy + 12xy + 9y2 + 9x2 - 24xy +16y2 = 25 25x2 + 25y2 = 25 25 (cos20 + sin20) = 25 25 (1) = 25 25 = 25 In order to solve this problem, I replaces cos and sin with x and y as I found them to be easier to work with and then I proceeded to factorize the equation. Once it was factorized I found that the rule that cos20 + sin20 equal 1came into play. At this point I inserted the cos and sin into parenthesis and placed the 125 outside the parenthesis in order to simplify the problem. By doing this sin and cos added up to one and I proved that (4sin0 + 3cos0)2 + (3sin0 - 4cos0)2 = 25.