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Trignometry

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6 Feb 2012 19:56:00 IST

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solution of triangles

Mathematics

in a triangle ABC if D is a point on side AB such that CD*CD=AD*BD. then prove that:sinAsinB<=sin(C/2)*sin(C/2)

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Nikhil Ulhas Karve

Cool goIITian

Joined: 18 Jan 2012

Posts: 31

7 Feb 2012 22:20:52 IST

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CD=b(sin(A))=a(sin(B))

I can write

Now,

AD=b.cos(A) and BD=a.cos(B)

Now its given That CD.CD=AD.BD

I get

and A+B=90.

which means C=90

Now as we know that sin of any angle is less than or equal to 1.

Now 2sin(A)cos(A)<=1

2sin(A)sin(B)<=1 (since C=90, cos(A)=sin(B))

i.e

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