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Ichar Nicole Heffernan
Grade 10
Trigonometry and Advanced Trigonometry Math Test
Level 1/2
1. Find the unknown length in the following diagram. Give your answer to 2 decimal
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
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
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
23. Prove the
following trigonometric identity:
18.1 km
(4sin0 +
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.