Question Number 124927 by bemath last updated on 07/Dec/20
Answered by liberty last updated on 07/Dec/20
$$\left(\mathrm{1}\right)\:\mathrm{sin}\:\mathrm{38}°\:=\:\frac{{x}+{y}}{\mathrm{8}}\:;\:{x}+{y}\:=\:\mathrm{8}.\mathrm{sin}\:\mathrm{38}° \\ $$$$\Rightarrow{x}+{y}\:=\:\mathrm{4}.\mathrm{9}\:{ft}\: \\ $$$$\left(\mathrm{2}\right)\:\mathrm{sin}\:\mathrm{52}°\:=\:\frac{{t}}{\mathrm{8}}\:;\:{t}\:=\:\mathrm{8}.\mathrm{sin}\:\mathrm{52}°\:=\:\mathrm{6}.\mathrm{3}\:{ft} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{tan}\:\mathrm{20}°\:=\:\frac{{x}}{{t}}\:;\:{x}\:=\:{t}.\mathrm{tan}\:\mathrm{20}°\:=\:\mathrm{2}.\mathrm{3} \\ $$$${then}\:{we}\:{get}\:{y}\:=\:\mathrm{4}.\mathrm{9}−\mathrm{2}.\mathrm{3}\:=\:\mathrm{2}.\mathrm{6}\:{ft} \\ $$
Commented by liberty last updated on 07/Dec/20
