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A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $\frac{13}{4}$. What are the angles of the triangle?

Here is the problem as I understand it, given that:
1. $\measuredangle ABC = 90^\circ \rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.

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Let $\measuredangle ACB = 2\alpha \rightarrow \measuredangle ACG = \alpha$
$\measuredangle BAC = 90 - 2\alpha \rightarrow \measuredangle GAC = 45 - \alpha$

From here we can construct the following system:

$$\begin{cases} 2R = \frac{r}{\tan \alpha} + \frac{r}{\tan(45 - \alpha)} \\ \frac{R}{r} = \frac{13}{4} \rightarrow R = \frac{13r}{4} \end{cases}$$ $$\downarrow \\ \frac{26r}{4} = r\left(\frac{1}{\tan \alpha} + \frac{1}{\tan(45 - \alpha)}\right) \quad / \div r \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos(45 - \alpha)}{\sin(45 - \alpha)} \quad / {\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha\\ \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos \alpha \cos 45 + \sin \alpha \sin 45}{\sin 45 \ cos \alpha - \sin \alpha \cos 45} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha } + \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \cdot \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \\ 6.5 = \frac{\cos^2 \alpha - \sin\alpha \cos \alpha + \cos \alpha \sin \alpha + \sin^2 \alpha}{\sin(\cos\alpha - \sin \alpha)} \quad / \sin^2\alpha + \cos ^2 \alpha = 1\\ 6.5 = \frac{1}{\sin\alpha (\cos\alpha - \sin \alpha)}$$

From this point on, I have no identity that I can think of which can be applied here to help me solve for $\alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.

I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.

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  • $\begingroup$ try rewriting your last equation as $\cos\alpha = \dfrac{1}{6.5\sin\alpha}+\sin\alpha$. What happens when you square both sides? $\endgroup$ – John Joy Jan 23 at 18:37
  • $\begingroup$ @JohnJoy I worked it out your way, and I get two possible solutions for an angle, with one of them being correct after disproving the other. However, I think that this amount of calculation is much beyond the level of this problem, and I suspect there's a simpler solution. Nevertheless, this works very well — thank you. $\endgroup$ – daedsidog Jan 23 at 18:58
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We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get $$\frac{a}{c}+\frac{c}{a}=\frac{169}{336}$$ from here you will get the quotient of $$\frac {a}{c}$$ and thus we can calculate the angles of triangle $$\Delta ABC$$

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  • $\begingroup$ What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example. $\endgroup$ – daedsidog Jan 23 at 19:10
  • $\begingroup$ Why not? They are the side lengths of the given triangle $\endgroup$ – Dr. Sonnhard Graubner Jan 23 at 19:11
  • $\begingroup$ I have all written what you need to solve the problem $\endgroup$ – Dr. Sonnhard Graubner Jan 23 at 19:11
  • $\begingroup$ You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$\frac{R}{r}=\frac{13}{4}$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 23 at 19:13
  • $\begingroup$ $2b \neq R$, but $b = 2R$. Typo on your part? $\endgroup$ – daedsidog Jan 23 at 19:16
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We have $$2R=\frac{r}{\tan(\alpha/2)}+\frac{r}{\tan(\beta/2)},$$ hence $$\frac{13}{2}=\frac{1}{\tan(\alpha/2)}+\frac{1}{\tan(\beta/2)}.$$ We know that $\alpha+\beta=\pi/2$, so $\beta/2=\pi/4-\alpha/2$. Let $t=\tan(\alpha/2)$. From the addition theorems for $\sin$ and $\cos$ derive that $$\tan(\pi/4-\alpha/2)=\frac{1-t}{1+t}.$$ Substitute and solve the quadratic equation for $t$.

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$$\dfrac4{13}\sin(45^\circ -\alpha+\alpha)=2\sin\alpha\sin(45^\circ-\alpha)$$

$$=\cos(45^\circ-2\alpha)-\cos45^\circ$$

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