Question Number 86932 by jagoll last updated on 01/Apr/20
Answered by mr W last updated on 01/Apr/20
$$\mathrm{1}. \\ $$$${RS}={RT} \\ $$$$\mathrm{5}\left({x}−\mathrm{3}\right)+\mathrm{6}=\mathrm{4}\left(\mathrm{2}{x}−\mathrm{6}\right) \\ $$$$\mathrm{15}=\mathrm{3}{x} \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{60}}=\frac{\mathrm{sin}\:\mathrm{50}}{\mathrm{250}}=\frac{\mathrm{sin}\:\left(\alpha+\mathrm{50}\right)}{{x}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{60}\:\mathrm{sin}\:\mathrm{50}}{\mathrm{250}} \\ $$$${x}=\frac{\mathrm{sin}\:\left(\alpha+\mathrm{50}\right)}{\mathrm{sin}\:\mathrm{50}}×\mathrm{250} \\ $$$$=\left(\frac{\mathrm{sin}\:\alpha}{\mathrm{tan}\:\mathrm{50}}+\mathrm{cos}\:\alpha\right)×\mathrm{250} \\ $$$$=\mathrm{284}.\mathrm{31} \\ $$