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प्रश्न
In ,Δ ABC with usual notations prove that
b2 = c2 +a2 - 2 ca cos B
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उत्तर
Consider that for Δ ABC , ∠B is in a standard position i.e. vertex B is at the origin and the side BC is along positive x-axis. As ∠B is an angle of a triangle ∴ ∠B can be acute or ∠B can be obtuse.
Using the Cartesian co-ordinate system in above figure.
we get B ≡(0,0) A ≡(c cos B , c sin B) and C ≡ (a,0)
Now consider l(CA) = b
∴ b2 = (a- c cos B)2 + (0-c sin B)2 , by distance formula
∴ b2 = a2 - 2 ac cos B + c2 cos2 B +c2 sin2 B
∴ b2 =a2 -2 ac cos B +c2 (sin2B + cos2B)
∴ b2 = a2 +c2 - 2 ac cos B
Hence proved.
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