Solve for "a" in terms of "r" when `theta` = 45degrees and v=3w : lnr=lna-ln(cos`theta` )-(v/w)ln(sec`theta` +tan`theta` ) Leave the answer in exact form.

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kingattaskus12's profile pic

kingattaskus12 | (Level 3) Adjunct Educator

Posted on

To solve for in `a` terms of `r,`

Substitute `theta = 45`  and `v = 3w`

`ln r = ln a - ln (cos 45) - ((3w)/w) ln(sec 45 + tan 45)`

Simplify,

`ln r = ln a - ln (sqrt(2)/2) - 3 ln (2/sqrt(2) + 1)`

Then use the Property of Natural Logarithm,

`ln r = ln a - ln (sqrt(2)/2) - ln (sqrt(2) + 1)^3`

Factor,

`ln r = ln a - [ln (sqrt(2)/2) + ln (sqrt(2) + 1)^3]`

Use again the Property of Natural Logarithm,

`lnr=ln(a /((sqrt(2)/2)(sqrt(2)+1)^3))`

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Raise each side by `e,`

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`e^(lnr)=e^(ln(a /((sqrt(2)/2)(sqrt(2)+1)^3)))`

` `  `r=a/((sqrt(2)/2)(sqrt(2)+1)^3)`

Expand `(sqrt(2)+1)^3`

`r=a/((sqrt(2)/2)(5 sqrt(2)+7))`

Multiply the denominator,

`r=a/(5+7 sqrt(2)/2)`

``Apply cross-multiplication,

`a = r (5 + 7 sqrt(2)/2)` 

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priwork's profile pic

priwork | High School Teacher | (Level 2) eNoter

Posted on

Given 

ln(r)=ln(a)-ln(cos )-(v/w)ln(sec +tan )

where θ = 45 degrees and v/w = 3

ln(a) =  ln(r) + ln(cosθ) +3ln(secθ+tanθ)

        = ln(rcosθ)+3ln (secθ+tanθ)

        =ln(rcosθ) +ln(secθ+tanθ)^3

        =ln(rcosθ).(secθ+tanθ)^3

`:.`  a= (rcosθ) (secθ+tanθ)^3

      =(r/√2)(1+√2)^3

      =(r/√2)(1+2√2+3√2+6)

      =(r/√2)(7+5√2)

    a  =(r/2) (7√2+10)    -Ans

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