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Ask community Community Discussion Question: solution of triangles
Reply Forum Index -> Trignometry originally posted here on IIT-JEE / AIEEE community   
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Arjun Blouria (0)

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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)
    
Nikhil Ulhas Karve (50)

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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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