Question Number 203269 by ahmetgg last updated on 13/Jan/24
Answered by esmaeil last updated on 13/Jan/24
$${tan}\mathrm{10}=\frac{{AH}}{{BH}} \\ $$$${tan}\mathrm{20}=\frac{{HC}}{{BH}}\rightarrow\frac{{HC}}{{AH}}\approx\mathrm{2}.\mathrm{0642} \\ $$$${tan}\mathrm{50}=\frac{{OH}}{{AH}} \\ $$$${tanx}=\frac{{OH}}{{CH}}\rightarrow\frac{{tan}\mathrm{50}}{{tanx}}\approx\mathrm{2}.\mathrm{0642}\rightarrow \\ $$$${x}={tan}^{−\mathrm{1}} \left(\frac{{tan}\mathrm{50}}{\mathrm{2}.\mathrm{0642}}\right)\approx\mathrm{30} \\ $$
Answered by mr W last updated on 14/Jan/24
$${AB}=\mathrm{1} \\ $$$${AC}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{110}°} \\ $$$${AH}=\frac{\mathrm{sin}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{40}°} \\ $$$${AC}={AH}\:\mathrm{cos}\:\mathrm{50}°+{AH}\:\mathrm{sin}\:\mathrm{50}°\:\mathrm{cot}\:{x} \\ $$$$\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{110}°}=\frac{\mathrm{sin}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{40}°}\:\left(\mathrm{cos}\:\mathrm{50}°+\mathrm{sin}\:\mathrm{50}°\:\mathrm{cot}\:{x}\right) \\ $$$$\mathrm{cot}\:{x}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{50}°}−\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{50}°} \\ $$$$\Rightarrow{x}=\mathrm{30}° \\ $$