Given a diagram like this,
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?)
Best Answer
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 $|OB|=\sqrt{26}$.