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Question Number 158035 by HongKing last updated on 30/Oct/21

Answered by mr W last updated on 30/Oct/21

say BC=1 ((BD)/(sin 9))=(1/(sin (3+9))) ⇒BD=((sin 9)/(sin 12)) ((BA)/(sin (9+45)))=(1/(sin (27+3+9+45))) ⇒BA=((sin 54)/(sin 84)) ((sin (x+27))/(sin x))=((BA)/(BD))=((sin 54)/(sin 84))×((sin 12)/(sin 9)) cos 27+((sin 27)/(tan x))=((sin 54)/(sin 84))×((sin 12)/(sin 9)) tan x=((sin 27)/(((sin 54)/(sin 84))×((sin 12)/(sin 9))−cos 27)) x=tan^(−1) (((sin 27)/(((sin 54)/(sin 84))×((sin 12)/(sin 9))−cos 27)))≈67.3°

$${say}\:{BC}=\mathrm{1} \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{9}}=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{3}+\mathrm{9}\right)}\: \\ $$$$\Rightarrow{BD}=\frac{\mathrm{sin}\:\mathrm{9}}{\mathrm{sin}\:\mathrm{12}} \\ $$$$\frac{{BA}}{\mathrm{sin}\:\left(\mathrm{9}+\mathrm{45}\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{27}+\mathrm{3}+\mathrm{9}+\mathrm{45}\right)} \\ $$$$\Rightarrow{BA}=\frac{\mathrm{sin}\:\mathrm{54}}{\mathrm{sin}\:\mathrm{84}} \\ $$$$\frac{\mathrm{sin}\:\left({x}+\mathrm{27}\right)}{\mathrm{sin}\:{x}}=\frac{{BA}}{{BD}}=\frac{\mathrm{sin}\:\mathrm{54}}{\mathrm{sin}\:\mathrm{84}}×\frac{\mathrm{sin}\:\mathrm{12}}{\mathrm{sin}\:\mathrm{9}} \\ $$$$\mathrm{cos}\:\mathrm{27}+\frac{\mathrm{sin}\:\mathrm{27}}{\mathrm{tan}\:{x}}=\frac{\mathrm{sin}\:\mathrm{54}}{\mathrm{sin}\:\mathrm{84}}×\frac{\mathrm{sin}\:\mathrm{12}}{\mathrm{sin}\:\mathrm{9}} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:\mathrm{27}}{\frac{\mathrm{sin}\:\mathrm{54}}{\mathrm{sin}\:\mathrm{84}}×\frac{\mathrm{sin}\:\mathrm{12}}{\mathrm{sin}\:\mathrm{9}}−\mathrm{cos}\:\mathrm{27}} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\mathrm{27}}{\frac{\mathrm{sin}\:\mathrm{54}}{\mathrm{sin}\:\mathrm{84}}×\frac{\mathrm{sin}\:\mathrm{12}}{\mathrm{sin}\:\mathrm{9}}−\mathrm{cos}\:\mathrm{27}}\right)\approx\mathrm{67}.\mathrm{3}° \\ $$

Commented by HongKing last updated on 31/Oct/21

cool my dear See thank you so much

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{ee}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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