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Trignometry

Scorching goIITian

Joined:	25 Jan 2009
Post:	246

17 Apr 2009 14:53:32 IST

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Sides and angles of a triangle

None

if p and q be the perpendiculars from the angular points A and B on any line passing through the vertex C of the triangle ABC, then prove that

a²p² + b²q² - 2abpq cos C = a²b² Sin² C

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Soumik

Blazing goIITian

Joined: 31 Jul 2008

Posts: 1266

18 Apr 2009 21:11:08 IST

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In some co-ordinate centre, select C as the origin....C=(0,0).

Frame B as (a,0), and A =(h,k).....Take the arbitary line through C as y=mx.

Drop perpendicular on BC from A.....of length k.

Thus 2apqbcosC= $\frac{2ahpq.\sqrt{h^2+k^2}}{\sqrt{h^2+k^2}}=2ahpq$

Similarly R.H.S = k²a².....

Thus a²p²= a²

b²q²= (h²+k²) $\frac{(am)^2}{m^2+1}$

So on simplifying both sides, we can easily see R.H.S=L.H.S.......

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