In triangle ABC, (a + B + C) × (a-b + C) = 3aC, then ∠ B=

In triangle ABC, (a + B + C) × (a-b + C) = 3aC, then ∠ B=


[(a+c)+b]×[(a+c)-b]=3ac
(a+c)²-b²=3ac
a²+c²-b²=ac
Then:
cosB=(a²+c²-b²)/(2ac)=1/2
Because:
0



To solve the equations 2010x-2011y = 2009, 2009x-2010y = 2008
Look at the question clearly and then answer ~
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2010x-2011y=2009 (1)
2009x-2010y=2008 (2)
subtract
x-y=1
Then 2010x-2010y = 2010 (3)
(3)-(2)
x=2
y=x-1=1



Use the appropriate method to solve the equations 2010x + 2011y = 20102011x + 2011y = 2011


Right minus left
So, x = 1
So y = 0