Question-212445

Question Number 212445 by ChantalYah last updated on 13/Oct/24

Commented by Frix last updated on 13/Oct/24

80cos A ? 150sin A =13 80cos A −150sin A =13 −170sin (A−tan^(−1) (8/(15))) =13 sin (A−tan^(−1) (8/(15))) =−((13)/(170)) A= { ((2nπ+tan^(−1) (8/(15)) −sin^(−1) ((13)/(170)))),(((2n+1)π+tan^(−1) (8/(15)) +sin^(−1) ((13)/(170)))) :} 80cos A +150sin A =13 170sin (A+tan^(−1) (8/(15))) =13 sin (A+tan^(−1) (8/(15))) =((13)/(170)) A= { ((2nπ−tan^(−1) (8/(15)) +sin^(−1) ((13)/(170)))),(((2n+1)π−tan^(−1) (8/(15)) −sin^(−1) ((13)/(170)))) :}

$$\mathrm{80cos}\:{A}\:?\:\mathrm{150sin}\:{A}\:=\mathrm{13} \\ $$$$ \\ $$$$\mathrm{80cos}\:{A}\:−\mathrm{150sin}\:{A}\:=\mathrm{13} \\ $$$$−\mathrm{170sin}\:\left({A}−\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\right)\:=\mathrm{13} \\ $$$$\mathrm{sin}\:\left({A}−\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\right)\:=−\frac{\mathrm{13}}{\mathrm{170}} \\ $$$${A}=\begin{cases}{\mathrm{2}{n}\pi+\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\:−\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{13}}{\mathrm{170}}}\\{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi+\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\:+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{13}}{\mathrm{170}}}\end{cases} \\ $$$$ \\ $$$$\mathrm{80cos}\:{A}\:+\mathrm{150sin}\:{A}\:=\mathrm{13} \\ $$$$\mathrm{170sin}\:\left({A}+\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\right)\:=\mathrm{13} \\ $$$$\mathrm{sin}\:\left({A}+\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\right)\:=\frac{\mathrm{13}}{\mathrm{170}} \\ $$$${A}=\begin{cases}{\mathrm{2}{n}\pi−\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\:+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{13}}{\mathrm{170}}}\\{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi−\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{8}}{\mathrm{15}}\:−\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{13}}{\mathrm{170}}}\end{cases} \\ $$

Answered by mr W last updated on 13/Oct/24

(8/( (√(8^2 +15^2 )))) cos A−((15)/( (√(8^2 +15^2 )))) sin A=((13)/(10(√(8^2 +15^2 )))) cos α cos A−sin α sin A=((13)/(170)) cos (α+A)=((13)/(170)) ⇒α+A=2kπ±cos^(−1) ((13)/(170)) ⇒A=2kπ±cos^(−1) ((13)/(170))−cos^(−1) (8/(17))

$$\frac{\mathrm{8}}{\:\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }}\:\mathrm{cos}\:{A}−\frac{\mathrm{15}}{\:\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }}\:\mathrm{sin}\:{A}=\frac{\mathrm{13}}{\mathrm{10}\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:{A}−\mathrm{sin}\:\alpha\:\mathrm{sin}\:{A}=\frac{\mathrm{13}}{\mathrm{170}} \\ $$$$\mathrm{cos}\:\left(\alpha+{A}\right)=\frac{\mathrm{13}}{\mathrm{170}} \\ $$$$\Rightarrow\alpha+{A}=\mathrm{2}{k}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{13}}{\mathrm{170}} \\ $$$$\Rightarrow{A}=\mathrm{2}{k}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{13}}{\mathrm{170}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{8}}{\mathrm{17}} \\ $$

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