maximum inscribed triangle

[Jedothek]

Please forgive me for posting a link, but it would be way too complicated to type it all out.
At

Maximum Area of Triangle - Optimization Problem with Solution

www.analyzemath.com

the site concludes ( see second step from end ) that

AB = AC * cos (45 degrees) = 2 r sqrt(2)
and CB = AC * sin (45 degrees) = 2 r sqrt(2)

This seems like an error to me; it seems to me that AB = CB = r sqrt(2).

Please either tell me I'm right or suggest where I may have gone wrong.

 

[Steven G]

In the picture below triangle ABC is inscribed inside a circle of center O and radius r. For a constant radius r of the circle, point B slides along the circle so that the area of ABC changes. Find the length of sides AB and CB so that the area of triangle ABC is maximum.

To OP. All I did was copy and paste.

If you think about it the right way the answer is obvious

 

[Dr.Peterson]

the site concludes ( see second step from end ) that

AB = AC * cos (45 degrees) = 2 r sqrt(2)
and CB = AC * sin (45 degrees) = 2 r sqrt(2)

This seems like an error to me; it seems to me that AB = CB = r sqrt(2).

Please either tell me I'm right or suggest where I may have gone wrong.

Yes, you're right. Since AC = 2r, and cos(45 deg) = sqrt(2)/2, AB = AC cos (45 deg) = 2 r * sqrt(2)/2 = r sqrt(2), not 2r sqrt(2). And likewise for CB.

 

[Jedothek]

Thank you!