# if p1,p2,p3 are respectively the perpendiculers from the vertices of a triangle to the opposite sides then prove that cosA/p1+cosB/p2+cosC/p3

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
8 years ago
$p1=2\bigtriangleup /a \\p2=2\bigtriangleup /b \\p3=2\bigtriangleup /c \\so \\acos(A)+bcos(B)+ccos(C)/2\bigtriangleup \\now \\acos(A)+bcos(B)+ccos(C)=2asin(B)sin(C)=2bsin(A)sin(C)=2csin(B)sin(A) \\sin(A)/a=sin(B)/b=sin(C)/c=2R //using\,these \\we\,get \\Answer=1/R$

Arun Kumar
IIT Delhi