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Question Number 84970 by Power last updated on 18/Mar/20

Answered by mr W last updated on 18/Mar/20

2x+3y=π ⇒x=(π/2)−((3y)/2) ((AB)/(sin (x+y)))=((BD)/(sin x)) ⇒((11)/(cos (y/2)))=(2/(cos ((3y)/2))) ⇒11=(2/(4 cos^2 (y/2)−3)) ⇒cos^2 (y/2)=((35)/(44)) ((AC)/(sin 2y))=((AB)/(sin y)) ⇒AC=2×AB cos y =2×AB(2 cos^2 (y/2)−1)=22(((70)/(44))−1)=13

$$\mathrm{2}{x}+\mathrm{3}{y}=\pi \\ $$$$\Rightarrow{x}=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}{y}}{\mathrm{2}} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\left({x}+{y}\right)}=\frac{{BD}}{\mathrm{sin}\:{x}} \\ $$$$\Rightarrow\frac{\mathrm{11}}{\mathrm{cos}\:\frac{{y}}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{cos}\:\frac{\mathrm{3}{y}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{11}=\frac{\mathrm{2}}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\frac{{y}}{\mathrm{2}}−\mathrm{3}} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \:\frac{{y}}{\mathrm{2}}=\frac{\mathrm{35}}{\mathrm{44}} \\ $$$$ \\ $$$$\frac{{AC}}{\mathrm{sin}\:\mathrm{2}{y}}=\frac{{AB}}{\mathrm{sin}\:{y}} \\ $$$$\Rightarrow{AC}=\mathrm{2}×{AB}\:\mathrm{cos}\:{y} \\ $$$$=\mathrm{2}×{AB}\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{{y}}{\mathrm{2}}−\mathrm{1}\right)=\mathrm{22}\left(\frac{\mathrm{70}}{\mathrm{44}}−\mathrm{1}\right)=\mathrm{13} \\ $$

Commented by Power last updated on 18/Mar/20

thanks

$$\mathrm{thanks}\: \\ $$

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