[Hình 10]

M

meou_a10

theo minh la
ta co cot A =cosA\sinA= ((b^2+c^2-a^2)\2bc)\(a\2R)= R*(b^2+c^2-a^2)\abc
tuong tu thi cos B = R*(a^2+c^2-b^2)\abc cosC=R*(a^2+b^2-c^2)\abc
ta se cot A +cotB+cotC=R\abc *(a^2+b^2+c^2)
ta co 4R\4abc= 1\4*(1\S)=1
vay ra dieu phai chung minh
 
T

tuongvy95

theo minh la:
Ta co: S=1/2absinC \RightarrowSinC=1/2ab
Tuong tu: sinA=1/2bc
sinA=1/2ac
cosA=(b^2+c^2-a^2)/2bc
cosB=(a^2+c^2-b^2)/2ac
cosC=(a^2+b^2-c^2)/2ab
cotA+cotB+cotC=cosA/sinA+cosB/sinB+cosC/sinC
=a^2+b^2+c^2:)>-
 
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