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?)

  • $\begingroup$ The angle OAB cannot be $45$ degrees. $\endgroup$ – Mark Sapir Jul 15 '20 at 15:01
  • 2
    $\begingroup$ 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

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 $|OB|=\sqrt{26}$.


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


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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.