Non-locus problem
Given that A=(-1,1), B=(2,0) and ΔABC is a right-angled triangle. Besides, AC = (sqrt5)
Now, from the calculation, we know that slope of AB = -1/3 and length of AB = sqrt10
So, BC = sqrt 5.
The problem is that, shouldn't tanB = slope of AB?
but tanB = 1, while slope of AB = -1/3
Hmm.. I'm confused...

Question

Non-locus problem
Given that A=(-1,1), B=(2,0) and ΔABC is a right-angled triangle. Besides, AC = (sqrt5)
Now, from the calculation, we know that slope of AB = -1/3 and length of AB = sqrt10
So, BC = sqrt 5. 
The problem is that, shouldn't tanB = slope of AB?
but tanB = 1, while slope of AB = -1/3
Hmm.. I'm confused...

anonymous · Answer

Not necessarily|dw:1334162896412:dw|

callisto · Answer

So, when is tanθ=slope of the line?

anonymous · Answer

I don't think it's necessary.

anonymous · Answer

tanB here is isn't the angle with the line and the x-axis. It's the angle of the right angle triangle.

anonymous · Answer

always but theta is what that matters

anonymous · Answer

|dw:1334163097066:dw|Slope of the line is tanθ, not tanB.