# Thread: Unable to get a step in the problem :(

1. ## Unable to get a step in the problem :(

Hey guys

The sides of triangle ABC are divided into segments of lengths a, b and x by the inscribed circle of centre O and radius r as shown. Find x in terms of a, b and r.

The first part of the answer by the teacher is as shown:

Let $\angle OAB = \alpha$ and $\angle OBA = \beta$
Therefore, $\angle OCA = 90 – (\alpha + \beta)$

So $tan (90 – (\alpha + \beta)) = \frac{x}{r}$
….

Could someone please explain to me how $tan (90 – (\alpha + \beta)) = \frac{x}{r}$? For some reason, all I seem to be able to get is $tan (90 – (\alpha + \beta)) = \frac{x}{r}$.

Thanx.

2. O is the intersection of the bisectors of the angles of triangle.

Then $\widehat{A}=2\widehat{OAB}=2\alpha, \ \widehat{B}=2\widehat{OBA}=2\beta$

$\widehat{OCA}=\frac{1}{2}\widehat{C}=\frac{1}{2}(1 80-2(\alpha+\beta))=90-(\alpha+\beta)$