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12:20
@SmokenSieEinBitteChebaHitBitswhen's the next lesson happening?
it's been a while..
 
2 hours later…
14:11
Hello @Buraian.
14:30
Hello! @NazmulHasanShipon
 
1 hour later…
15:54
It's good to see you @Buraian. I'd always be grateful for that you helped me previously.
Can you help me about a math problem? @Buraian
11
A: How to calculate the coordinates of a triangle's orthocentre?

najayaz Here $A\equiv (x_1,y_1)$,$B\equiv (x_2,y_2)$,$C\equiv (x_3,y_3)$. I'll use the usual notation for $a,b,c,R$ and $A,B,C$. Using simple trigonometry, $BP=c\cos B$, $PC=b\cos C\implies\dfrac{BP}{PC}=\dfrac{c\cos B}{b\cos C}$ $$\implies P_x= \dfrac{x_2b\cos C+x_3c\cos B}{b\cos C+c\cos B}= \dfrac{x...

Why not
type it out ill check in a bit
15:56
Can you tell me how he got AH=2RcosA and HP=2RcosBcosC ?
@Buraian okay
no idea on that one
 
2 hours later…
17:54
@Buraian, but he said after using some trigonometry.
18:10
he hasn't defined thevariables only
what on earth is R
 
2 hours later…
19:45
@Buraian, would you please look for me in your coordinate geometry books if there is the proof of the formula presented in the question?

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