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Question Number 46814 by peter frank last updated on 31/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18

cosθ=((OP^2 +OQ^2 −PQ^2 )/(2×OP×OQ))     =(((a^2 +b^2 )+(c^2 +d^2 )−{(a−c)^2 +(b−d)^2 })/(2×(√(a^2 +b^2 )) ×(√(c^2 +d^2 )) ))   =((ac+bd)/((√(a^2 +b^2 )) ×(√(c^2 +d^2 )) ))

$${cos}\theta=\frac{{OP}^{\mathrm{2}} +{OQ}^{\mathrm{2}} −{PQ}^{\mathrm{2}} }{\mathrm{2}×{OP}×{OQ}} \\ $$$$\:\:\:=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)−\left\{\left({a}−{c}\right)^{\mathrm{2}} +\left({b}−{d}\right)^{\mathrm{2}} \right\}}{\mathrm{2}×\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:×\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }\:} \\ $$$$\:=\frac{{ac}+{bd}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:×\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }\:} \\ $$

Commented by peter frank last updated on 01/Nov/18

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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