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The Brainliest Answer!
2015-06-21T18:35:17+05:30
R =  \sqrt{ a^{2} + b^{2} +2abcos(theta)}

 \sqrt{10}P = \sqrt{ (2P)^{2} + ( \sqrt{2}P) ^{2} +2(2P)( \sqrt{2}P)cos(theta) }

On solving we get θ= 45°
2 5 2