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Question-56165




Question Number 56165 by Tawa1 last updated on 11/Mar/19
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19
A=6cos36^0 i+6sin36^o j  B=−7i  A+B=(6cos36^o −7)i+6sin36^0 j  A−B=(6cos36+7)i+6sin36j  ∣A+B∣^∣ =(√((6cos36^0 −7)^2 +(6sin36^o )^2 ))  =(√(6^2 +7^2 −2×6×7cos36^o ))   ∣A−B∣=(√(6^2 +7^2 +2×6×7cos36_ ^o ))
$${A}=\mathrm{6}{cos}\mathrm{36}^{\mathrm{0}} {i}+\mathrm{6}{sin}\mathrm{36}^{{o}} {j} \\ $$$${B}=−\mathrm{7}{i} \\ $$$${A}+{B}=\left(\mathrm{6}{cos}\mathrm{36}^{{o}} −\mathrm{7}\right){i}+\mathrm{6}{sin}\mathrm{36}^{\mathrm{0}} {j} \\ $$$${A}−{B}=\left(\mathrm{6}{cos}\mathrm{36}+\mathrm{7}\right){i}+\mathrm{6}{sin}\mathrm{36}{j} \\ $$$$\mid{A}+{B}\mid^{\mid} =\sqrt{\left(\mathrm{6}{cos}\mathrm{36}^{\mathrm{0}} −\mathrm{7}\right)^{\mathrm{2}} +\left(\mathrm{6}{sin}\mathrm{36}^{{o}} \right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{2}×\mathrm{6}×\mathrm{7}{cos}\mathrm{36}^{{o}} }\: \\ $$$$\mid{A}−{B}\mid=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} +\mathrm{2}×\mathrm{6}×\mathrm{7}{cos}\mathrm{36}_{} ^{{o}} }\: \\ $$
Commented by Tawa1 last updated on 11/Mar/19
God bless you sir. I appreciate
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$

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