Help !! How can I find acute Angle theta??


by krautkramer
Tags: acute, angle, theta
krautkramer
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#1
Aug12-11, 02:41 AM
P: 25
How to find acute angle theta ,when tan 63 Degree = cot theta??Could anyone help me to find answer of this query.I need steps too as I am sooooo poor in mactematicuos .Final Answer of theta should be in degrees

Thanks in advance
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krautkramer
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#2
Aug12-11, 03:01 AM
P: 25
1. The problem statement, all variables and given/known data

Help !! How can I find acute Angle theta??
How to find acute angle theta ,when tan 63 Degree = cot theta??Could anyone help me to find answer of this query.I need steps too as I am sooooo poor in mactematicuos .Final Answer of theta should be in degrees

Thanks in advance

2. Relevant equations

I don't know,If I know these things,then I would have been the head of 'NASA' :)

3. The attempt at a solution
NascentOxygen
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#3
Aug12-11, 03:13 AM
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P: 4,703
First step, instead of cot write it as tan, you are probably more comfortable with tan.

If you don't know what cot is, ask google.

PeterO
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#4
Aug12-11, 03:16 AM
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P: 2,316

Help !! How can I find acute Angle theta??


Quote Quote by krautkramer View Post
1. The problem statement, all variables and given/known data

Help !! How can I find acute Angle theta??
How to find acute angle theta ,when tan 63 Degree = cot theta??Could anyone help me to find answer of this query.I need steps too as I am sooooo poor in mactematicuos .Final Answer of theta should be in degrees

Thanks in advance

2. Relevant equations

I don't know,If I know these things,then I would have been the head of 'NASA' :)

3. The attempt at a solution
tan, is short for tangent
cot is short for co-tangent - which is short for complementary-tangent.

In the same way sin and cos are short for sine and complementary-sine

In the case of sine and cosine, the sine of an angle, and the complementary-sine of the complementary angle are equal in value.

eg, 41 degrees and 49 degrees are complementary [they add up to 90]

compare sin 41 to cos 49 - or if you prefer sin 49 to cos 41.

The relationship between tan and cot is the same.
krautkramer
krautkramer is offline
#5
Aug12-11, 03:49 AM
P: 25
Hi,thank you friends for the help,but it is not adequate to solve my problem.

I know simple mathematics like sin/cos=tan and cos/sin=cot, but not able to find a solution for my critical problem.

I will repeat the query once again...Hope you can help me....

Determine the acute angle θ when tan 63 = cot θ.
Attached Thumbnails
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krautkramer
krautkramer is offline
#6
Aug12-11, 04:17 AM
P: 25
Nobody here to help me out...
thebiggerbang
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#7
Aug12-11, 04:20 AM
P: 68
It's simple.

Use the given equation to obtain
tan(63)=cot[itex]\theta[/itex]
arccot(tan(63))=[itex]\theta[/itex]

Solving, [itex]\theta[/itex]=27
krautkramer
krautkramer is offline
#8
Aug12-11, 04:23 AM
P: 25
Quote Quote by thebiggerbang View Post
It's simple.

Use the given equation to obtain
tan(63)=cot[itex]\theta[/itex]
arccot(tan(63))=[itex]\theta[/itex]

Solving, [itex]\theta[/itex]=27
hey, what is this 'arccot'??How I can find the answer 27 degree?? by calculator or any other means...
PeterO
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#9
Aug12-11, 04:30 AM
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P: 2,316
Quote Quote by krautkramer View Post
Hi,thank you friends for the help,but it is not adequate to solve my problem.

I know simple mathematics like sin/cos=tan and cos/sin=cot, but not able to find a solution for my critical problem.

I will repeat the query once again...Hope you can help me....

Determine the acute angle θ when tan 63 = cot θ.
Are you allowed to use a calculator when you do this question?
krautkramer
krautkramer is offline
#10
Aug12-11, 04:33 AM
P: 25
Quote Quote by thebiggerbang View Post
It's simple.

Use the given equation to obtain
tan(63)=cot[itex]\theta[/itex]
arccot(tan(63))=[itex]\theta[/itex]

Solving, [itex]\theta[/itex]=27
Yes of course...You are right and the answer is 27 dgree..it was simple for you...But for me,it is a herculian task...still I don't know how to find it by using windows scientific calculator
krautkramer
krautkramer is offline
#11
Aug12-11, 04:39 AM
P: 25
Quote Quote by PeterO View Post
Are you allowed to use a calculator when you do this question?
Yes of course...I think so...Otherwise no one will pass the final exam...
PeterO
PeterO is offline
#12
Aug12-11, 04:48 AM
HW Helper
P: 2,316
Quote Quote by krautkramer View Post
Hi,thank you friends for the help,but it is not adequate to solve my problem.

I know simple mathematics like sin/cos=tan and cos/sin=cot, but not able to find a solution for my critical problem.

I will repeat the query once again...Hope you can help me....

Determine the acute angle θ when tan 63 = cot θ.
Try the following:

Write down an angle - eg 17
Using your calculator:
Take tan 17
use 1/x, the reciprocal, on the answers
now do inverse tan and write down that answer.

repeat starting with an angle different to 17 degrees.

Note anything?
PeterO
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#13
Aug12-11, 04:51 AM
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Quote Quote by krautkramer View Post
Yes of course...I think so...Otherwise no one will pass the final exam...
In that case, try each of the options in the cot function.

If your calculator doesn't have a cot function, use the tan function followed by the 1/x function.

[you best take the tan 63 to start with and write that answer down for reference.
krautkramer
krautkramer is offline
#14
Aug12-11, 05:05 AM
P: 25
Quote Quote by PeterO View Post
In that case, try each of the options in the cot function.

If your calculator doesn't have a cot function, use the tan function followed by the 1/x function.

[you best take the tan 63 to start with and write that answer down for reference.
Hoooorahhh, Got it..Today,I learned lot of mactamactics...Thanks and 1000 hugs to sonic generation and other friends....My examination body should allow us to use internet during exam.

http://answers.yahoo.com/question/in...1201353AAYQeXG


Determine the acute angle θ when tan 63 = cot θ.

Tan63 =Cot θ
Ie, 1.96261= Cot θ

Arccot(1.96261)=Cot θ

Ie, Cot θ=1.96261

Tan θ=1/1.96262

So, θ=Tan-1(1/1.96262)

Ans=27
PeterO
PeterO is offline
#15
Aug12-11, 05:36 AM
HW Helper
P: 2,316
Quote Quote by krautkramer View Post
Hoooorahhh, Got it..Today,I learned lot of mactamactics...Thanks and 1000 hugs to sonic generation and other friends....My examination body should allow us to use internet during exam.

http://answers.yahoo.com/question/in...1201353AAYQeXG


Determine the acute angle θ when tan 63 = cot θ.

Tan63 =Cot θ
Ie, 1.96261= Cot θ

Arccot(1.96261)=Cot θ

Ie, Cot θ=1.96261

Tan θ=1/1.96262

So, θ=Tan-1(1/1.96262)

Ans=27
The information / expression you were really holding out for - and we were holding back on is


Tan θ = Cot (90 - θ) or Cot θ = tan (90 - θ)

Similarly

Sin θ = Cos (90 - θ) or Cos θ = Sin (90 - θ)

so Tan 63 = Cot (90 - 63) = Cot 27

That is the complementary angle stuff I was trying to lead you to.
asmani
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#16
Aug12-11, 07:06 AM
P: 84
Notice that tan(θ)=cot(90-θ) for any θ.
HallsofIvy
HallsofIvy is offline
#17
Aug12-11, 08:19 AM
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Thanks
PF Gold
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This was posted in both "Precalculus Homework" and "General Mathematics" so I am merging the two threads.


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