# Thread: Could someone look over some problems?

1. ## Could someone look over some problems?

I don't have anyone that can help me with my review questions except for the few things i've been able to google. I am practicing review questions for an upcoming exam but the professor refused to give us an answer guide to check our work. I thought the point of a review question was to practice them at home and still be able to see if you got the correct answer but apparently he doesnt see it that way.

Would anyone have the time to possibly look over the 10 problems i have completed so far and let me know if i am on the right path?

I am not looking for someone to give me answers, i provide the answer i have gotten and am also showing my work so you can see how i got my answer. I really just want someone to let me know if im doing things correctly so i can fix any errors.

Not sure where i should have posted this but all the problems are pre-calculus ones.

I am new to this site so i suppose to pm me and i can email you the scanned copies of what i have done? Thank you all for your help =)

2. Originally Posted by FailingTrig
Not sure where i should have posted this but all the problems are pre-calculus ones.

I am new to this site so i suppose to pm me and i can email you the scanned copies of what i have done? Thank you all for your help =)
Post them in a thread with your attempts, you will get a variety of help this way.

3. ## Here are my attempts:

4. Q1.jpg looks right.

Q3-1.jpg:
In part b, a nitpick but it looks like you're multiplying a power of i by a number:
$i^2 \cdot 103$
when you meant multiplying exponents:
$i^{2 \cdot 103}$
Since $i^4 = 1$, I would have done the problem like this:
\begin{aligned}
i^{206} &= i^{204} \cdot i^2 \\
&= 1 \cdot (-1) \\
&= -1
\end{aligned}

in part e, while you did the work correctly, usually you need to write complex answers in standard form, and your answer
$\frac{-3 + 29i}{25}$
isn't technically in standard form. Break up the fraction:
$-\frac{3}{25} + \frac{29}{25}i$

5. Q4-1.jpg:

(A) is right.

(B) another nitpick about standard form; we don't put the i in front of the fraction (although I notice that it was written that way in part (C)!). The answer should be written as
$\frac{5}{2} - \frac{5 \sqrt{3}}{2}i$.

(C) I would have done it this way:
I recognize that if cos θ = -1/2 and sin θ = -√(3)/2 then θ = 4π/3.
\begin{aligned}
\left[ 5 \left( -\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) \right]^3 &= \left[ 5 cis \left( \frac{4\pi}{3} \right) \right]^3 \\
&= 125 cis 4\pi \\
&= 125
\end{aligned}

Q5.jpg:

If I were grading this, I would appreciate you showing what i is in trig form as your first step:
$i = 0 + 1i = cos \left( \frac{\pi}{2} \right) + i \, sin \left( \frac{\pi}{2} \right)$
Also, again you need to write in correct standard form:
$\frac{(\sqrt{3} + i)}{2} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$
$\frac{(-\sqrt{3} + i)}{2} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i$

6. It gets confusing since the professor writes it one way and the book, like yourself places the i in a different area. I'm afraid of writing it the "correct" way for fear of him marking it wrong because it isnt written *his* way. Thank you very much for taking the time to look over these questions for me and informing me of other ways to do them and the correct forms as well. I doubt i would have been told this useful information by anyone else. =)

7. Have i done this problem correctly?

http://i953.photobucket.com/albums/a...incomplete.png < problem itself

My response:
sin(60) /184.5 = sin(180-60-x)/123
sin(120-x) = 123 sin(60) / 184
x = 120 - Arcsin ( 123 sin(60) / 184 )
x ~ 84.62°

8. Originally Posted by FailingTrig
Have i done this problem correctly?

http://i953.photobucket.com/albums/a...incomplete.png < problem itself

My response:
sin(60) /184.5 = sin(180-60-x)/123
sin(120-x) = 123 sin(60) / 184
x = 120 - Arcsin ( 123 sin(60) / 184 )
x ~ 84.62°
Some rounding error, I think, maybe because the "184" in the last two lines is missing the ".5"? I got x ≈ 84.74°, but anyway, that's not the angle the problem is asking -- angle x is angle RPQ, and this is the angle that they want (the angle is marked with "**") in the diagram (click below):

This angle is easy to find.

For part (B), you drop a perpendicular from R to PQ, as in this diagram (click below):

$\triangle PRS$ is a right triangle, and PR, its hypotenuse, is 184.5, so you should be able to find the length RS.

9. Originally Posted by eumyang
Some rounding error, I think, maybe because the "184" in the last two lines is missing the ".5"? I got x ≈ 84.74°, but anyway, that's not the angle the problem is asking -- angle x is angle RPQ, and this is the angle that they want (the angle is marked with "**") in the diagram (click below):

This angle is easy to find.

For part (B), you drop a perpendicular from R to PQ, as in this diagram (click below):

$\triangle PRS$ is a right triangle, and PR, its hypotenuse, is 184.5, so you should be able to find the length RS.

Law of sines:
a/sinA = b/sinB =c/sinC
sin(60)/(184.5) = sin(x)/(123)
sin(x) = sin(60) * 123 / 184.5
x = 33.080 degrees

180 - (33.080 + 60) = 86.920 degrees
90 - 86.920 = 3.080 degrees <~~~ Angle of Tilt

sin(x) = O/H
sin(86.920) = R/184.5
R = 184.5 * sin(86.920)
R = 169.146 ft.

maybe?

10. Close!

Originally Posted by FailingTrig
Law of sines:
a/sinA = b/sinB =c/sinC
sin(60)/(184.5) = sin(x)/(123)
sin(x) = sin(60) * 123 / 184.5
x = 33.080 degrees
I don't know how you got that answer. I got about 35.26°.

sin(x) = O/H
sin(86.920) = R/184.5
R = 184.5 * sin(86.920)
R = 169.146 ft.
Ignoring the fact that the angle is wrong, I don't know how you got this number either. When I entered 184.5 * sin(86.920°) in my calculator I got 184.23.