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Given a diagram like this, enter image description here

Where $O$ is the center and $OA = \sqrt{50}$, $AB = 6$, and $BC = 2$. The question was to find the length of $OB$. $\angle ABC = 90^o$

What I've done is so far:

I made the triangle $ABC$ and named $\angle BAC = \alpha$ . By trigonometry, I have the values for $\sin{\alpha}$ and $\cos{\alpha}$. I get $\cos{\alpha}=\frac{6}{\sqrt{40}}$.

Then I made the triangle $OCA$ and named $\angle OAB = \beta$ so $\angle OAC = \alpha + \beta$. By using the cosinus rule, I have $\cos(\alpha + \beta) = \frac{1}{\sqrt{5}}$.

Using the formula, $\cos(\alpha + \beta) = \cos{\alpha}.\cos{\beta} - \sin{\alpha}.\sin{\beta}$ and making $\sin{\beta} = \sqrt{1 -\cos^2{\beta}}$ I finally get that $\cos{\angle OAB} = \frac{1}{\sqrt{2}}$.

Finally, by using the cosinus rule on the triangle $AOB$ I get $OB = \sqrt{26}$.

My only problem is this takes me way too long! I am interested in a quicker way to do this (i.e. I now know that $\angle OAB = 45^o$ from trigonometry, but is there a quicker way to recognize it?)

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  • $\begingroup$ The angle OAB cannot be $45$ degrees. $\endgroup$ – Mark Sapir Jul 15 '20 at 15:01
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    $\begingroup$ Are we also given that $\angle ABC$ is right angle? $\endgroup$ – Jaap Scherphuis Jul 15 '20 at 15:09
  • $\begingroup$ @JaapScherphuis yeah! I forgot, sorry. $\endgroup$ – aco Jul 15 '20 at 15:41
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Assuming $\angle ABC=90^o$ is given.

You can get there slightly quicker:
By Pythagoras, $|AC|=\sqrt{40}$.
$OAC$ is isosceles, with $|OA|=|OC|=\sqrt{50}$.
You can then immediately get $\cos(\angle OAC)=\frac{|AC|/2}{|OA|} = \frac{\sqrt{40}/2}{\sqrt{50}}= \frac{1}{\sqrt{5}}$.
I don't yet see a way to shortcut the rest.

You could do it completely differently, by algebra. Use a coordinate system, centred on $B$, and let $O$ be the point $(x,y)$. Then we get two equations from the fact that $|OA|=|OC|=\sqrt{50}$.

$$x^2+(6-y)^2=50\\ (2-x)^2+y^2=50$$

There are fairly easily solved to give $y=1$, $x=-5$, from which you get $|OB|=\sqrt{26}$.

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Refer to the figure:

$\hspace{4cm}$enter image description here

From the right triangle $ACD$: $CD=\sqrt{AD^2-AC^2}=4\sqrt{10}$.

From similarity of right triangles $ABC$ and $CDE$: $$\frac{CE}{AB}=\frac{CD}{AC}\Rightarrow CE=12\\ DE=\sqrt{CD^2-CE^2}=4=BF\\ BE=CE-BC=12-2=10=DF=AF$$ Hence, $\angle DAF=45^\circ=\angle OAB$, indeed.

Finally, from the cosine theorem for $\triangle AOB$: $$\begin{align}BO&=\sqrt{AO^2+AB^2-2\cdot AO\cdot AB\cdot \cos \angle OAB}=\\ &=\sqrt{50+36-2\cdot \sqrt{50}\cdot 6\cdot \frac1{\sqrt2}}=\\ &=\sqrt{26}.\end{align}$$

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enter image description here

A slight variation of the solution

Note that $R$ is circumradius of $\triangle ADC$,

\begin{align} |CD|&=2R\sin\alpha=2\sqrt5 ,\\ |BD|&=\sqrt{|CD|^2-a^2}=4 ,\\ |AD|&=c+BD=10 . \end{align}

By the Stewart’s Theorem for $\triangle AOD$,

\begin{align} |OD|^2\cdot c+|OA|^2\cdot |BD| -|AD|\cdot(|OB|^2+c\cdot |BD|) &=0 ,\\ |AD|\cdot ( R^2 -|OB|^2-c\cdot |BD|) &=0 , \end{align}

\begin{align} |OB|^2&= R^2-c\cdot |BD| \\ &=50-6\cdot4 =26 . \end{align}

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