Question

Does anyone have the time to check about 5-7 problems for me? I already have my answers. This is new material (sin/cos/tan) and I would like to make sure I am doing it right. Thank you ! I will post the first 2 questions below!

Answer 1

I circled my answers

Answer 2

good job

Answer 3

I hope you haven't solve these questions using calculator.

Answer 4

10*cos(36)


18*sin(35)

Answer 5

Did i get it wrong?! :(

Answer 6

look good to me, just practice setting up the cos(a) and sin(a) ratios

Answer 7

You aren't wrong.

Answer 8

tangent uses the sides...

inverse tangent of that ratio ..

Answer 9

and use cosine for the last

Answer 10

Dan, is that for the next set i put (26 & 27) or the first set (21-22)?

Answer 11

all look good... what are you unsure about?

Answer 12

Trigonometry is just new to me, so I am having you all check my answers to make sure my mind is on the right track

how about this, I got B for #28 >

Answer 13

ok, for 28, what did you do

Answer 14

you can just type the thought process if you want

Answer 15

For 28, I am given the hypotenuse and opposite so I used sin. method
8/14 (opp/hyp) = 0.571
the I did sin^-1 (0.571)= 34.8 (and then I have to round up to 35)

Answer 16

right, after a few practice, using sin, cos, tan, just becomes like a pattern, look opposite side and hyp for sin, not really thinking opposite and hypotenuse...

Answer 17

29. B
30. B

these are the last 2 I need check, you know, just for assuring purposes lol

Answer 18

Sorry I had so many questinos, I am just very prone to making simple mistakes and getting things wrong, even when I knw how to do it xD

Answer 19

inverse tangent, and inverse sin of those ratios

Answer 20

here, some examples

Answer 21

here is a reference sheet for things

Answer 22

Alrighty Thank you for checking them!!

do you think you have time to walk me through one problem that I do not know how to work out that involves manipulating a formula? (it is the only one I cannot figure out on my whole assignment)

Answer 23

sure

Answer 24

alright, i am pretty good at formulas, but this one i have no idea

Answer 25

ok, that is just some algebra , moving things around

Answer 26

im gonna right down the steps as we go if thats all right (for future reference0

Answer 27

\[\large A = 2\pi*r^2 + 2\pi*r*h\]

Answer 28

ok,..notice that is the 2* area of a circle + circumference*height, or the surface area of a cylinder

you want so solve for pi = something

Answer 29

I have to end up with an equation that solves for \[\pi = ?\] not numerical values, but variables

Answer 30

2 steps is easiest,

factor the common pi from both of those terms

\[\large A = 2\pi*r^2 + 2\pi*r*h = \pi*[2r^2 + 2rh]\]

Answer 31

then just divide by all that in the quantity

Answer 32

left with pi = A/[ ]

Answer 33

\[\pi = \frac{ a }{ 2r^2 + 2rh }\] ?

Answer 34

yep, factor the pi out, then divide by all the [ ] stuff, leaving pi by itself

Answer 35

ooh ok, see I was trying to factor 2pi r or something like that, idk what I was doing, but your way makes perfect sense!

Answer 36

Thank you so much for taking the time to help me! I really feel like I learned! Thank you!

Answer 37

yes, the 2 and the r, are common factors there, but in the end you want pi by itself, just more work your way

no prob, anytime