9.3 #33

I don't know, most of you probably already understand how to do this problem, but personally it tripped me up, so here's a solution.

You know the magnitude of v is 1/2 and the direction angle is 250degrees. It helped me to draw a right triangle in the third quadrant of a set of axes, with a hypotenuse of 1/2. Since the direction angle is 250 degrees, this means the hypotenuse makes an angle of 70degrees with the negative x-axis. I then labeled the horizontal leg "x" and the vertical leg "y" and said that cos(70)=x/(1/2) or cos(70)=2x and that sin(70)=y/(1/2) or sin(70)=2y. So x=cos70/2 and y=sin70/2. You can use a calculator to approximate these values, but don't forget to make each one negative, since they are in the third quadrant! The component form of the vector is thus <-.17101, -.469846>.

#17-19

I want to ask about number 17 and 19
thanks

Problem 10 on the Pre Test

I have been working all night with what energy I have, on this problem but have gotten no where! Please help

Solution to 35

Ok, this requires using the identity we proved in class:

[p*cis(a)]*[q*cis(b)]=(pq)*cis(a+b)

Remember that cis(a) means cos(a)+i*sin(a)

so, for polar form, apply the above formula:
12cis(11*pi/12)*(7/2)*cis(pi/4)=42*cis(7*pi/6)

For rectangular form, multiply this out:
42*(cos(7pi/6)+i*sin(7pi/6))=42*(-sqrt(3)/2-i/2)=-21*sqrt(3)-21i

There you go!

#35

Can someone please post a solution to #35 for the homework

9.1 33

Is the answer in the book a typo? I got the correct polar coordinates, but the rectangular coordinates the book gives do not seem to work.

9.1 # 19

I'm having trouble understanding this one. It asks you to sketch a graph of all complex numbers z that would fulfill the equation absvalue(z-1)=10. I know this is a circle with radius 10, but I'm having a hard time determining the center. The book hints that "1" corresponds to the complex point (1,0) and asks how far apart z and 1 are. I think this would mean that z is one unit to the left of z-1, but I don't know how to figure out how far apart z and (1,0) are. Basically... the book makes the circle's center (1,0) and I'm wondering if anyone understands why. I'm probably overthinking it...

new assignment sheet

hey guys did we ever get a new assignment sheet? i checked the website and i don't see a new one up, so does that mean we have no hw due for tomorrow?

Pretest Questions

i'm basically done with the pretest except for these 4 questions. can someone please explain them to me?

10 and 14. i don't even know where to start. any suggestions?

11. i started by finding the diagonal on the bottom face: sqrt(2x^2), if the sides of the cube=x and labeling the angle ABC theta.
tan(theta)=x/sqrt(2x^2)
--> theta=tan^-1(x/sqrt(2x^2))
where do i go from here to get real numbers or is there a better way?

15. I have the diagram labeled but i'm not sure how to find AB. should i use the right triangle ABD (if BD is the median to AC)?

thanks.

10 on pretest

does anyone have a suggestion as to how to find the angle of the arc, or a better idea? thanks

# 2 on back of Assn. Sheet

Is it supposed to be angle uOz/u? Otherwise, what does it have to do with division?

#15

In order to solve triangles it is best to draw a simple equilateral triangle. Label the angles of the triangle A, B, and C labeling clockwise starting where ever. The opposite side of the angle will be the corresponding side to the angle A with a, B with b and so on. For this specific problem c is 14 and a is 10. We also know that angle B will be 130 degrees. In order to find the last remaining side, use the law of cosines and solve for c. Now we can solve for andle a, using the law of cosines. After finding this angle we can add it to the given angle and subtract the resulting sum from 180; calculating angle C.

homework

For the review question homework, i want to ask about number 18 and 35 because i didnt know how to set up the diagram. Thanks

Answer to #18

There is a diagram for the problem below. Knowing that the angle of the obtuse triangle we're concerned with is 107°, just use the law of cosines to solve for the third side (x). I think it comes out to be like 92? I'm doing this all from memory by the way so I'm not sure this is entirely accurate.

#18 Review probs.

#18) How do u find the length of the wire? Is the angle that the wire makes with the hill 163 degrees? And is the wire going up the hill or down the hill?

Numbers 35/37

For number thirty five could someone just help me out with solving the problem once I have the man down to the right of the building and the woman on the top right edge? And for 37 is the fastest way ot solve to fine the lengths of the partially visible triangle using law of cosines or is there a better way?

# 35

For the picture, start with a 448 ft foot vertical line. At the top, this opens into a 152 degree angle, which is the womens 62 degree angle of elevation plus an additional 90 degree angle, because angle of elevation is from a horizontal line, and shes on a vert line. At the base of the 448 ft segment, there is a 25 degree angle opening into the plane. This is 90-65, 65 is his angle of elevation. This makes a small 3 degree angle at the top. After i set up this diagram, I kept getting incorrect answers. Any critiques would be appreciated.

ch. 7 review

can anyone explain 35 to me? i think i set up my diagram wrong.

thanks

Diagram for #18 (Hill Problem)

In case the image didn't work...
Draw a slanted line. This will be the hill. Next draw a vertical line going upward from a point on the hill. This will be your flag pole. Next draw a segmant that is perpendicular to your flagpole and intersects the hill. This is your horizontal. The angle between the horizontal and the hill is 17 in this problem. Draw a segmant connecting the top of the flagpole and the hill somewhere below the base. The length of this segment is the unknown. You now have a triangle and can use the law of cosines to determine the unknown.

#18

Can anyone explain how to set up number 18 from the review problems? specifically, i don't understand how to find the angle, since it gives the angle the hill makes with the horizontal (which would be outside the triangle) but not the angle the hill makes with the wire.
i also need help with the diagram for number 35 i think because i keep getting the wrong answers.

problems concerning hills

i dont understand these, there is one on the hw. how do i set it up

Ch. Review

Is the chapter review mandatory or is it just for extra credit?

Solving parallelograms w/ diagonals

i am having problems figuring this out. i am aware that the magic formula is 1/2*ab*sin C, but how do i get this when, for instance, the diagonlas are given as 18 and 24, with angle 30 deg. like on the groupwork. i heard the answer was 108. how does this make sense?

7.3.47

I really like this problem so if anyone would like to hear a brief explanation, here it is.

1. The statue's height is 6.3m (this is the fisrt side of your triangle[side a]) From the base of the triangle to some point of=n the hill side is 36m (second side b).The final side of your triangle is unknown and will be found (side c).

2. Using side c as your hypoteneus, draw a triangle attached to triangle abc so that we will be able to solve for side c and the complementary angles of the 10 degree angle from the top of the statue to the point on the side of the hill.

3. Solve triangle abc, and the right triangle that shares the hypoteneus of abc.

4. Subtract the 10 degree hill side angle and the angle of the right triangle attached to the 10 degree angle, from 90.

5. The resulting angle is the angle from the horizontal to the point on the side of the hill (5.27 degrees).

7.3.48 (Sorry this is late, I didn't have internet)

Yeah sorry about the lateness.

So, #48

This works best with a diagram, so here goes:
B___c____A
-\ ~~~~~~/
--\~~~~~/
-a-\~~~/-b
----\~/
-----C

angle B=58 degrees
a=40
b=36
OK, so there's no way to go through this without dealing with angle C or side C, so I think it's easiest to go with the angle. To do that, you have to find angle A. So, according to the law of sines, sin(58)/36=sin(A)/40, so A=asin(40*sin(58)/36)=70.4 degrees. Therefore angle C=180-70.4-58=51.6 degrees.

So, then draw an altitude from A to a. Call the intersection of the altitude and a point D. Ok, so, to find the length of the altitude (the height of the triangle, just take the sine of angle C and multiply it by the length of a, which should come out to 31.3 feet. Then multiply that by the base, b, which comes out to 1253.9 square feet.

Hurray!

#13 on Chapter 8 Test

If f(x)=cos(x), find Df(x) and rewrite in the form (_____)cos(x) - (_____)sin(x). I just don't know what to do with the "h" in the Df(x) function, after you get it to ((cos(x)cos(h)-sin(x)sin)(h)-cos(x))/h.

Solving for angle on TI-89 in limited range

Even though you can use the inverse trig functions to find an angle, sometimes i'd rather just have my calculator to solve an equation with the variable inside a trig function. This works fine except that you get all the possible solutions, and if you want to find the solution in say [0,pi] you have to figure it out yourself. When I ask it to find an answer in a limited range it gives me all the answers anyway.
Any ideas on how to make it give you an answer in a given range?

Alternate form of Law of Cosines

The alternate form of the law of cosines is

cos(c)= (a^2+b^2-c^2)/(2ab)

this is helpful when solving equations to simply figure out the angle rather than having to solve the whole equation and then solving for the cos of the angle.

#'s 35 and 39

I don't understand what # 35 is asking. How are you supposed to find where the tracking station is and then where the satellite will be at 2:05?
And for #39 I'm so confused. How am I supposed to fin the length of AD or BE with out the angle DAC or EBC? I kno that the length of arc DE is pi but then what?

review packet

hey guys, i know that Mr. Karafiol said that he emailed the review packet but i didnt recieve it, can somebody please repost it up in the yahoo group. thanks.
melissa

7.1 #63

I drew a right triangle, facing right. The left vertical side is the lamp. The bottom side is the to length distance from the women to the shadow, and then an additional 4 feet of shadow. You drop a perpendicular line where the ten and four meet. You have two lengths and an angle, so you can find the hypotenuse of the small triangle. When you have the ratio of hypotenuse to base in the small triangle, you can figure it out for the larger, similar triangle.

7.1 #63

can anyone explain how to set up number 63 from 7.1, specifically how to find all the side lengths of the two triangles?
(I know this is from a slightly old assignment, but i'm catching up from missing school all last week, so if anyone could look back and explain it that would be great)

Trouble with 30

How do you draw a diagram for this problem?

Review Questions

Can someone please email me the review questions to my gmail account? i can't check my yahoo right now.

i.am.monica.91@gmail.com

Thanks

7.2->28

I'm havving trouble drawing a diagram/picture for this problem, wut does it mean when it says the boats "diverge by 38*"
if someone can show me wut their equation was, im sure i will be able to figure out the drawing if thasts easier

Solution to 7.2 #34

This is the diagram for #34.

a. Since angle DAC=55 degrees and angle DAB=70 degrees, angle CAB=70-55=15 degrees. CB^2=AC^2+AB^2-2*AC*AB*cos(15)=180^2+400^2-2*180*400*cos(15), CB=230.88

b. Measure angle ACB. 400^2=180^2+230.88^2-2*180*230.88*cos(angle ACB), angle ACB=153.36degrees. The angle of turn = 180-153.36= 26.64degrees.

review questions

did anyone in 4th period get the review packet?

Did Mr. Karaiol ever pass out the Chap. 7 review problems in class? Because he just e-mailed us corrections for the sheet, but i definitely don't have the sheet/don't remember him passing it out. But it could just be me.

7.2#34

I'm having trouble drawing thios figure. I drew it such that i had two triangles with 1 side and 2 angles but don't know how to use this information.
Also, did we have review problems?

i know the assignment that this question was on was due a while back, but can someone explain question 96 in section 8.4?

last chapter

i liked that last chapter, proofs are fun!

nsolve()

If an equation cannot be solved using Solve() or cSolve(), nsolve() may give you a correct solution. For example the equation (x+1)^(x+2)=0 will return false, but if you plug it into nsolve(), it will return -1.

Pretest

On the pretest, I dont understand and dont know how to get the last question. Thanks.
melissa

Number 6 on pretest

The domain of sin -1 (ln x) is________________?
Does anyone have any suggestions? I know the domain of sin^-1 (x) is the range of sin(x) which is [-1.1]. However, how does the ln(x) fit in?

#16 on the pretest

does anyone have an easy way to do this problem...

# 3 on Pretest

Does anyone have an idea for # 3 on the pretest. I am pretty stuck. Thanks