# How to establish arc actions from angles in between secant and also tangents (without trigonometry)

Given a circle, a factor $H$ outside the circle, sectors $\overline{HE}$ and also $\overline{HT}$ tangent to the circle at $E$ and also $T$, specifically, and also factors $I$ and also $G$ on the circle such that $I$, $G$, and also $H$ are collinear (all as revealed over), recognizing the actions of $\angle EHG$ and also $\angle GHT$ (call them $\alpha$ and also $\beta$, specifically, for ease) establishes the actions of each of the 4 arcs on the circle.

Is it feasible to calculate the actions of the small arcs $EI$, $IT$, $TG$, and also $GE$ in regards to $\alpha$ and also $\beta$ without making use of trigonometry?

I'm within one, so I wish it motivates your reasoning. Standard geometry was a great deal of years earlier. The theories I bear in mind are $\stackrel{\frown}{EGT} = 2\angle EIT$, $\beta=(\stackrel{\frown}{IT} - \stackrel{\frown}{TG})/2$, and also $ET=\pi-\alpha-\beta$, which you can see by attracting distances from E and also T and also the line from H to the facility of the circle. After that if $\angle EIG=\gamma$ we have $EG=2\gamma, GT=\pi-\alpha-\beta-2\gamma, TI=\pi-\alpha+\beta-2\gamma, IE=2\gamma+2\alpha$

*With * trig, I get that the arcs have action $\theta \pm \alpha$ and also $\pi - \theta \pm \beta$, where $$\cos\theta = \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} = \frac{\cos\beta-\cos\alpha}{1-\cos\left(\alpha+\beta\right)}$$

By the "far arc / near arc" principle, $\stackrel{\frown}{EI}-\stackrel{\frown}{EG} = 2\alpha$, so we can write $\stackrel{\frown}{EI} = \theta+\alpha$ and also $\stackrel{\frown}{EG}=\theta-\alpha$.

Allow is go logical ...

Let the circle to have distance $1$, with its facility, $O$, at the beginning ; and also allow $H$ lie on the favorable $x$ - axis. In quadrilateral $OEHT$, angles at $E$ and also $T$ are appropriate angles, and also the angle at $H$ has action $\alpha+\beta$, to make sure that the angle at $O$ has action $2 \gamma := \pi-(\alpha+\beta)$. Hence, $\angle HOE = \gamma$ and also $|OH| = \sec\gamma$. Walking around the circle by arcs of $\theta \pm \alpha$ from $E$ obtains us to $I$ and also $G$, and also we establish works with for these factors, creating "$\rm{cis}\cdot$" for "$(\cos\cdot,\sin\cdot)$" :

$$I = \rm{cis}\left(\gamma + ( \theta + \alpha )\right) = \rm{cis}\left(\gamma+\alpha+\theta\right)$$ $$G = \rm{cis}\left(\gamma-(\theta-\alpha)\right) = \rm{cis}\left(\gamma+\alpha-\theta\right)$$

Since $H$, $G$, and also $I$ are collinear, we have to have equivalent inclines for sectors $HI$ and also $HG$ :

$$\frac{ \sin(\gamma+\alpha+\theta) }{ \cos(\gamma+\alpha+\theta)-\sec\gamma } = \frac{ \sin(\gamma+\alpha-\theta) }{ \cos(\gamma+\alpha-\theta)-\sec\gamma }$$

Therefore,

$$\sin(\gamma+\alpha+\theta)\left(\cos(\gamma+\alpha-\theta)-\sec\gamma\right) = \sin(\gamma+\alpha-\theta) \left( \cos(\gamma+\alpha+\theta)-\sec\gamma\right) $$

$$\begin{eqnarray*} &&\sin(\gamma+\alpha+\theta)\cos(\gamma+\alpha-\theta)- \sin(\gamma+\alpha-\theta) \cos(\gamma+\alpha+\theta) \\ &=& \sec{\gamma} \; \left( \sin(\gamma+\alpha+\theta) -\sin(\gamma+\alpha-\theta) \right) \\ \sin 2\theta &=& 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right) \\ 2 \sin\theta \cos\theta &=& 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right) \end{eqnarray*}$$ If $\theta\ne 0$ and also $\theta \ne \pi$ [ * ], after that we can terminate $\sin\theta$ and also coating - up : $$\begin{eqnarray*} \cos\theta &=& \frac{\cos\left(\gamma+\alpha\right)}{\cos\gamma} = \frac{\cos\frac{\pi+\alpha-\beta}{2}}{\cos\frac{\pi-(\alpha+\beta)}{2}}= \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} \\ \end{eqnarray*}$$

There is absolutely a means to get to this outcome without making use of works with and also slopes (I obtained careless), yet if there is a means to arrive "without making use of trigonometry", I'm not seeing it.

**Edit. **
I'll simply mention that, if $F$ is the foot of the vertical from the circle is facility to sector $HI$, after that

$$\frac{|OF|}{|OE|}=\frac{|OH| \sin\angle FHO}{|OH|\sin\angle EHO}=\frac{ \sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}=\cos\left(\pi-\theta\right)$$

Also, there are factors $P$ and also $Q$ on the circle, with $P$ on $\stackrel{\frown}{IE}$ and also $G$ on $\stackrel{\frown}{QE}$, such that $\stackrel{\frown}{IP} = \stackrel{\frown}{GQ} = \alpha$, whence $\stackrel{\frown}{PE}=\stackrel{\frown}{QE}=\theta$. So, all the parts of the last relationship show up in the number someplace. I have not looked hard sufficient to establish if it is feasible to "see" the relationship geometrically. **End edit. **

[ * ] If $\theta = 0$, after that $\alpha$ itself have to be $0$ (to make sure that $\theta-\alpha$ is a non - adverse arc action), with $I$ and also $G$ accompanying $E$. Furthermore, $\theta = \pi$ indicates $\beta = 0$, with $I$ and also $G$ accompanying $T$. Evaluation of these instances is uncomplicated, with the very same formula resulting as in the basic instance.

There is no remedy without inverted trigonometric features, due to the fact that any kind of primary geometric argument would certainly, for an offered value of the angle in between the tangent lines, compute all 4 arc sizes as straight features of $\alpha$. This would indicate that, as the (secant) line via $H$ is revolved at consistent angular rate, it purges arclength on the circle at a consistent price. Yet this is difficult due to the fact that the range from $H$ to the junction factor, $|HG|$, is not constant. It is not tough to make this argument strenuous making use of calculus.

If the arcs, in the order detailed in the inquiry, are represented $A, B, b, a$, after that primary geometry does share $(A-a)$, $(B-b)$, and also $(A+B+b+a)$ as straight features of $\alpha, \beta$ and also $1$, as in the various other remedies. I assume this set of 3 formulas in 4 unknowns is what the OP suggested by "not adequate equations" to address every little thing totally geometrically.

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