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Question Number 170445 by cortano1 last updated on 24/May/22

Solve ((sin x)/(sin (x+5°))) = (1/(2cos 55°))

$$\:\:\:{Solve}\:\:\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\mathrm{5}°\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{55}°} \\ $$

Answered by thfchristopher last updated on 24/May/22

((sin x)/(sin (x+5°)))=(1/(2cos 55°)) ⇒2sin xcos 55°=sin (x+5°) cos 55° =cos (60°−5°) =cos 60°cos 5°+sin 60°sin 5° =((cos 5°+(√3)sin 5°)/2) ∴2sin xcos 55°=sin (x+5°) ⇒sin xcos 5°+(√3)sin xsin 5°=sin xcos 5°+cos xsin 5° ⇒tan x=(1/( (√3))) x=180n+30° n=0,1,2,....

$$\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\mathrm{5}°\right)}=\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{55}°} \\ $$$$\Rightarrow\mathrm{2sin}\:{x}\mathrm{cos}\:\mathrm{55}°=\mathrm{sin}\:\left({x}+\mathrm{5}°\right) \\ $$$$\mathrm{cos}\:\mathrm{55}° \\ $$$$=\mathrm{cos}\:\left(\mathrm{60}°−\mathrm{5}°\right) \\ $$$$=\mathrm{cos}\:\mathrm{60}°\mathrm{cos}\:\mathrm{5}°+\mathrm{sin}\:\mathrm{60}°\mathrm{sin}\:\mathrm{5}° \\ $$$$=\frac{\mathrm{cos}\:\mathrm{5}°+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{5}°}{\mathrm{2}} \\ $$$$\therefore\mathrm{2sin}\:{x}\mathrm{cos}\:\mathrm{55}°=\mathrm{sin}\:\left({x}+\mathrm{5}°\right) \\ $$$$\Rightarrow\cancel{\mathrm{sin}\:{x}\mathrm{cos}\:\mathrm{5}°}+\sqrt{\mathrm{3}}\mathrm{sin}\:{x}\cancel{\mathrm{sin}\:\mathrm{5}°}=\cancel{\mathrm{sin}\:{x}\mathrm{cos}\:\mathrm{5}°}+\mathrm{cos}\:{x}\cancel{\mathrm{sin}\:\mathrm{5}°} \\ $$$$\Rightarrow\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${x}=\mathrm{180}{n}+\mathrm{30}°\:\:\:{n}=\mathrm{0},\mathrm{1},\mathrm{2},.... \\ $$

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