Question Number 218853 by Lekhraj last updated on 16/Apr/25

Answered by mr W last updated on 17/Apr/25

Commented by mr W last updated on 17/Apr/25

$${A}_{{blue}} =\frac{\mathrm{3}×\mathrm{5}\:\mathrm{sin}\:\left(\mathrm{60}°+\alpha\right)}{\mathrm{2}}=\frac{\mathrm{3}×\mathrm{5}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{3}\left(\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{4}} \\ $$
Answered by Spillover last updated on 17/Apr/25

Answered by Spillover last updated on 17/Apr/25

Answered by Spillover last updated on 17/Apr/25

Answered by Spillover last updated on 17/Apr/25
