How do I get T out of this equation? (trig)

[madflame]

I'm doing some Statics and I need to find the tension in a rope. I did some math and ended up with this equation:

T * sin [ cos^-1 ( 10.93 / T ) ] = 165.9

I try to put this in my calculator (Ti-89 Titanium) and solve using the 'zeros' function, but it gives me some pretty weird results.

So can someone please tell me how I get T out of all that? ANY help is appreciated. I come up with similar equations for other problems that I'm working. Thanks.

 

[galactus]

There's a little identity which when known will help solve this.

\(\displaystyle \L\\Tsin(cos^{-1}(\frac{10.93}{T}))=165.9\)

Identity:

\(\displaystyle \L\\sin(cos^{-1}(\frac{a}{b}))=\frac{\sqrt{b^{2}-a^{2}}}{b}\)

So, you're problem is:

\(\displaystyle \L\\sin(cos^{-1}(\frac{10.93}{T}))=\frac{\sqrt{T^{2}-119.4649}}{T}\)

So, we have:

\(\displaystyle \L\\T\cdot\frac{\sqrt{T^{2}-119.4649}}{T}=165.9\)

\(\displaystyle \L\\\sqrt{T^{2}-119.4649}=165.9\)

\(\displaystyle \L\\T^{2}-119.4649=27522.81\)

\(\displaystyle \L\\T=\sqrt{27642.2749}=166.26\)

 

[madflame]

Thank you!

 

[soroban]

Hello, madflame!

I'll try to explain Galactus' identity . . .

\(\displaystyle T\cdot\sin\left[\cos^{-1}\left(\frac{10.93}{T}\right)\right]\: = \:165.9\)

An inverse trig function is an angle.

That is: \(\displaystyle \,\cos^{-1}\left(\frac{10.93}{T}\right)\) is some angle \(\displaystyle \theta.\)

So we have: \(\displaystyle \,\theta\:=\:\cos^{-1}\left(\frac{10.93}{T}\right)\)

Take the cosine of both sides: \(\displaystyle \,\cos\theta \:=\:\frac{10.93}{T}\)

So \(\displaystyle \theta\) is an angle in a right triangle with: \(\displaystyle adj\,=\,10.93\) and \(\displaystyle hyp\,=\,T\)

                  *
           T   *  |
            *     |
         * θ      |
      * - - - - - *
          10.93

Using Pythagorus, we find that: \(\displaystyle \,opp\,=\,\sqrt{T^2\,-\,10.93^2}\)

Then: \(\displaystyle \,\sin\theta\:=\:\frac{opp}{hyp}\:=\:\frac{\sqrt{T^2\,-\,10.93^2}}{T}\)


Substitute into the original equation: \(\displaystyle \:T\cdot\frac{\sqrt{T^2\,-\,10.93^2}}{T} \:=\:165.9\)

Then: \(\displaystyle \:\sqrt{T^2\,-\,10.93^2} \:=\:165.9\)

. . . . . . \(\displaystyle T^2\,-\,10.93^2\:=\:165.9^2\)

. . . . . . . . . . . .\(\displaystyle T^2\:=\:165.9^2\,-\,10.93^2\)

. . . . . . . . . . . . \(\displaystyle T \;= \;\sqrt{165.9^2\,-\,10.93^2} \:\approx\:165.54\)