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If the segments joining the points `A(a , b)a n d\ B(c , d)` subtends an angle `theta` at the origin, prove that : `costheta=(a c+b d)/sqrt((a^2+b^2)(c^2+d^2))`

Text Solution

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Here `(AB)^(2) = (a - c)^(2) + (b -d)^(2)`
`(OA)^(2) = (a -0)^(2) + (b -0)^(2) = a^(2) + b^(2)`
and `(OB)^(2) = c^(2) + d^(2)`

Now in `Delta AOB`,
`cos theta = ((OA)^(2) + (OB)^(2) - (AB)^(2))/(2OA xx OB)`
`= (a^(2) + b^(2) + c^(2) + d^(2) - [(a -c)^(2) + (b -d)^(2)])/(2 sqrt(a^(2) + b^(2)) sqrt(c^(2) + d^(2)))`
`= (ac + bd)/(sqrt((a^(2) + b^(2)) (c^(2) + d^(2))))`
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