Solve 1-betacotbeta=frac{h_{0}R}{k}

$$1 - \beta \cot \beta = \frac { h _ { 0 } R } { k }$$

Solve for R

$$\left\{\begin{matrix}R=-\frac{k\left(\beta \cos(\beta )-\sin(\beta )\right)}{h_{0}\sin(\beta )}\text{, }&h_{0}\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\pi n_{1}\text{ and }k\neq 0\\R\in \mathrm{R}\text{, }&\beta \cos(\beta )-\sin(\beta )=0\text{ and }h_{0}=0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\pi n_{1}\text{ and }k\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by k.

$$k-\beta \cot(\beta )k=h_{0}R$$

Swap sides so that all variable terms are on the left hand side.

$$h_{0}R=k-\beta \cot(\beta )k$$

The equation is in standard form.

$$h_{0}R=-k\beta \cot(\beta )+k$$

Divide both sides by h_{0}.

$$\frac{h_{0}R}{h_{0}}=\frac{k\left(-\beta \cos(\beta )+\sin(\beta )\right)}{\sin(\beta )h_{0}}$$

Dividing by h_{0} undoes the multiplication by h_{0}.

$$R=\frac{k\left(-\beta \cos(\beta )+\sin(\beta )\right)}{\sin(\beta )h_{0}}$$

Divide \frac{k\left(\sin(\beta )-\beta \cos(\beta )\right)}{\sin(\beta )} by h_{0}.

$$R=\frac{k\left(-\beta \cot(\beta )+1\right)}{h_{0}}$$

Solve for h_0

$$\left\{\begin{matrix}h_{0}=-\frac{k\left(\beta \cos(\beta )-\sin(\beta )\right)}{R\sin(\beta )}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\pi n_{1}\text{ and }R\neq 0\text{ and }k\neq 0\\h_{0}\in \mathrm{R}\text{, }&\beta \cos(\beta )-\sin(\beta )=0\text{ and }R=0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\beta =\pi n_{1}\text{ and }k\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by k.

$$k-\beta \cot(\beta )k=h_{0}R$$

Swap sides so that all variable terms are on the left hand side.

$$h_{0}R=k-\beta \cot(\beta )k$$

The equation is in standard form.

$$Rh_{0}=-k\beta \cot(\beta )+k$$

Divide both sides by R.

$$\frac{Rh_{0}}{R}=\frac{k\left(-\beta \cos(\beta )+\sin(\beta )\right)}{\sin(\beta )R}$$

Dividing by R undoes the multiplication by R.

$$h_{0}=\frac{k\left(-\beta \cos(\beta )+\sin(\beta )\right)}{\sin(\beta )R}$$

Divide \frac{k\left(\sin(\beta )-\beta \cos(\beta )\right)}{\sin(\beta )} by R.

$$h_{0}=\frac{k\left(-\beta \cot(\beta )+1\right)}{R}$$