IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

viperissiva

for an acute angle triangle prove that (sinA/A)+(sinB/B)+(sinC/C)>6/pie

plz help fast...................................

surya_kant_2005

viperissiva

y isn't any1 replyin...............no1 knows the answer kya???????

viperissiva

damn it.....................all r silent.............y is this capped when no1 has answered this question!!!!!!!!!!

elessar_iitkgp

As A,B,C are acute angled,
sinA<A

(sinA)/A <1
Similarly
(sinB)/B <1
(sinC)/C <1
Adding these
(sinA)/A + (sinB)/B + (sinC)/C < 3

Also
2>

2/

>1

6/

>3

6/

>(sinA)/A + (sinB)/B + (sinC)/C

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