Question

solve the equation 2 tan c-3=3 tan C-4 algebraically for all values of c in the interval 0≤C<360

Answer 1

@celecity plz helppp

Answer 2

@AccessDenied

Answer 3

@satellite73

Answer 4

Are these the same variable C or is there a difference between lowercase and uppercase c/C's?

Answer 5

its same?

Answer 6

Okay. Have you made any attempts so far? Or not sure how to begin?

\( 2 \tan C - 3 = 3 \tan C - 4\) \( 0 \le C < 360\)

Answer 7

do u add 3 on both sideS?

Answer 8

That would be fine.
\( 2 \tan C = 3 \tan C - 1\)
Then we can move 3tan C to the left side.

Answer 9

im gona do 2tanc -3=3tanc-4?

Answer 10

From 2 tan C = 3 tan C - 1
We can subtract the entire 3 tan C from both sides to group it on the left side. Can you see how that works out?
2 tan C = 3 tan C - 1
2 tan C - 3 tan C = 3 tan C - 1 - 3 tan C
(2 - 3) tan C = 0 - 1

Answer 11

ugh im totally confused on what ur doin can u guide me step by step plz?

Answer 12

So, I'll start from the initial equation:
\( 2 \tan C - 3 = 3 \tan C - 4 \)
You were correct to say to add 3 to both sides. I will colorize the additions made.
\( 2 \tan C - 3 \color{green}{ + 3} = 3 \tan C - 4 \color{Green}{+ 3} \)
\( 2 \tan C = 3 \tan C - 1 \)
This is where we were moments ago. The next step is as if we had this equation with u in place of tan C.
\( 2 u = 3 u - 1 \)
You would subtract 3u from both sides to solve for u, yes?

Answer 13

\( 2u = 3u - 1 \) ==> \( 2u \color{Green}{- 3u} = 3u - 1 \color{green}{- 3u} \) ==> \(-u = -1\)

\( 2 \tan C = 3 \tan C - 1 \)
==> \( 2 \tan C \color{green}{- 3 \tan C} = \cancel{3 \tan C} - 1 \color{green}{\cancel{- 3 \tan C}} \)
==> \( (2 - 3) \tan C = -1 \)
==> \( - \tan C = -1 \)

Answer 14

but y r u choosing u over c?

Answer 15

u is replacing tan C, to make a point about how to solve.
u = tan C would make this look more like a simple single-variable equation like "2u - 5 = 6"

Answer 16

It's not significant to the problem itself, but rather to make it more evident to you that the solution comes from simply trying to solve for the "tan C" term first.

Answer 17

oh like x? 3x-2=0 lets say

Answer 18

Yes. If you wanted to solve for that x, you would take a similar route. It is the same with the entire "tan C" part and u.

\( 2x = 3x - 1 \) \( 2\tan C = 3 \tan C - 1 \)
\(2x \color{green}{- 3x} = 3x - 1 \color{Green}{- 3x}\) \(2 \tan C \color{green}{- 3 \tan C} = 3 \tan C - 1 \color{green}{- 3 \tan C} \)
\((2-3) x = -1 \) \((2 - 3) \tan C = -1 \)
\(-1 x = -1\) \(-1 \tan C = -1 \)

Answer 19

ok gt it now im up to the step wher i got -tanc=-1

Answer 20

what do i do afta?

Answer 21

you'd want "tan C" alone, so the coefficient needs to go to the right side.
we'd multiply both sides by -1 (or divide, both are same operation...)
\( -x = -1\) \( -\tan C = -1 \)
\(-1 * -x = -1 * -1 \) \( -1 * -\tan C = -1 * -1 \)

Answer 22

which gets you tan C = 1.

Answer 23

ok i got tht,.

Answer 24

den?

Answer 25

Last part is finding what values of C will get you a value for tangent of 1.
If you know the unit circle, this is one to find on that. tan C = (sin C)/(cos C), so where sin C = cos C.
tan C = 1 0 <= C < 360

Answer 26

or a calculator may also tell you the value here:
tan C = 1
C = tan^(-1) (1)
but because it is restricted to -180 to 180, you will miss any values 180< C< 360.

Answer 27

I got -45?

Answer 28

is that tan^(-1) (1) or tan^(-1) (-1) ?

Answer 29

by tan^(-1) (1) I mean:
\( \tan^{-1} \left(1\right) \)

Answer 30

i stl gt -45

Answer 31

weird. that looks like the value for \( \tan^{-1} \left(-1 \right) \)

Answer 32

i type \[\tan^(-1) \]

Answer 33

tan −1 (−1)

Answer 34

i type 2ndtan(-1)

Answer 35

Do you agree we ended up with this equation?
tan C = 1 <-- positive one?

Answer 36

yes i gt tht

Answer 37

we take inverse tangent of both sides:
\( \tan^{-1} \left( \tan C \right) = \tan^{-1} \left( 1 \right) \)
\( C = \tan^{-1} \left(1 \right) \)
Apologies if the nonformatted answer was confusing. But you should type 2nd tan(1) for positive 45 degrees.

Answer 38

ohh now i got 45!

Answer 39

yep. :)
the only other thing to note is that this was only the solutions in -90 < C < 90, the range of inverse tangent.
for 90 < C < 270 , the other solution is 45 + 180 = 225 degrees. this is because tangent has a period of 180 degrees (pi radians).

Answer 40

so for C:
tan (C + 180) = tan C

Answer 41

how did u no to do tht tho?

Answer 42

we wanted to find all the values of C over this interval: 0 <= C < 360, a given.
tangent is a periodic function with period 180. we restrict its domain to [-90, 90] to make the inverse function, a function with a single value.

when we took tan ^(-1) (1), we obtained the value for C in the domain of tangent [-90, 90]. but in general, any value: C + 180n will give the same value as C because tangent is periodic.
tan (C + 180) = tan C
(a) We found C = 45 as one solution. What about 45 + 180 n ?
(b) Adding 180 to 45, C = 180+45 = 225 is also in our interval 0 <= C < 360.
(c) Adding 180 again to 225, 405 is NOT in our interval of values for C.

Answer 43

if you had written C = 45 + 180n from the start (and did this every time), the problem becomes "how many integer values of n will get us a value of C in the interval: 0 <= C < 360 ? "
C = 45 + 180(0) = 45
C = 45 + 180(1) = 225

Answer 44

ohh i get now thanks a lot ur awesome!

Answer 45

glad to help! :)

Answer 46

ru a teacher by any chance?

Answer 47

nope, i am a high school senior. :D

Answer 48

wow nice u had the patience to explain it to me. wt u state do u liv in?

Answer 49

Illinois. :)
Yes, I have a lot of patience for Math. (I recall spending 6 hours or more helping someone on OS with geometry one night, heh).

Answer 50

wow thts crzy i have another ques last one for 2night tho lol

Answer 51

find algebraically, the measure of the obtuse angle to the nearest degree, tht satisfies 5 csc theta=8

Answer 52

\( 5 \csc \theta = 8 \)
I would recommend converting reciprocal functions like cosecant and secant to sines and cosines. It makes them easier to deal with.
\( 5 \dfrac{1}{\sin \theta} = 8 \)

Answer 53

ok then?

Answer 54

Multiply both sides by sin theta.
\( 5 = 8 \sin \theta \)
and then get rid of coefficient on sin theta by dividing it off both sides:
\( \dfrac{5}{8} = \dfrac{\cancel{8}}{\cancel{8}} \sin \theta \)
that makes sense so far?

Answer 55

multiply the 8 by theta? and wht else/?

Answer 56

Uhh, we won't need to multiply by anything from here. We have sin theta = a fraction. Now we can take inverse sine of both sides.

Answer 57

no im stil up to the 1st step

Answer 58

Oh, this was a division by theta:
\( 5 = 8 \sin \theta \)
\( \dfrac{5}{8} = \sin \theta \)

Answer 59

ERR, division by eight ***

Answer 60

how r u gtting 5=8sin theata?

Answer 61

oh i'm sorry, i see what you were questioning.
\( 5 \csc \theta = 8 \)
\( 5 \dfrac{1}{\sin \theta} = 8 \) <-- multiplying both sides by \( \sin \theta \)

\( 5 \dfrac{1}{\sin \theta } \times \sin \theta = 8 \times \sin \theta \)

Answer 62

the sin theta divided by sin theta cancels on the left, leaving 5.
\( 5 = 8 \sin \theta \)

Answer 63

what happened to the one?

Answer 64

\( \dfrac{1}{\sin \theta} \times \sin \theta = \dfrac{1}{\sin \theta} \times \dfrac{\sin \theta}{1} \)
dividing by 1 does not change value.
You multiply fractions straight across, that sounds familiar?

Answer 65

right so now im up to 5=8sintheata

Answer 66

yup.
\( 5 = 8 \sin \theta \)
did it make sense when I said solve for \( \sin \theta \) by removing the coefficient?

( like 5 = 8 x )

Answer 67

so sin theats = 5/8?

Answer 68

correct :)

Answer 69

k then wha?

Answer 70

so from there, we just need to use inverse sine on both sides. calculating sin^(-1) (5/8) should give us the value of theta for -180 < theta < 180

Answer 71

38.7?

Answer 72

y u subtracting 180 tho?

Answer 73

woops,
the range is -90 < theta < 90 for arcsin >.<
keep mixing it up thinking about radians

Answer 74

lol

Answer 75

but then recall this part:
"obtuse angle"
obtuse angles... how big is the obtuse angle from geometry?

Answer 76

180?

Answer 77

got it thnks!

Answer 78

sooo
we're looking for an angle for \(\theta \) greater than 180 and less than 360. (definition of obtuse)

Answer 79

thnks a lot! gn :D

Answer 80

basically, we found the angle 38.7. but thats much lower than 180.
We need to use the angle 180+38.7 this time, because that angle lies in the correctinterval

Answer 81

gnight!