Question Number 209590 by SonGoku last updated on 15/Jul/24
Commented by SonGoku last updated on 15/Jul/24
$$\mathrm{Help}\:-\:\mathrm{me}! \\ $$
Answered by mr W last updated on 16/Jul/24
$$\frac{{A}_{\Delta{big}} }{{A}_{\Delta{small}} }=\frac{\left(\mathrm{1}+\mathrm{20}\right)×\left(\mathrm{0}.\mathrm{85}+\mathrm{17}\right)}{\mathrm{1}×\mathrm{0}.\mathrm{85}} \\ $$$${A}_{\Delta{small}} =\frac{\sqrt{\left(\mathrm{1}+\mathrm{0}.\mathrm{85}+\mathrm{1}.\mathrm{095}\right)\left(−\mathrm{1}+\mathrm{0}.\mathrm{85}+\mathrm{1}.\mathrm{095}\right)\left(\mathrm{1}−\mathrm{0}.\mathrm{85}+\mathrm{1}.\mathrm{095}\right)\left(\mathrm{1}+\mathrm{0}.\mathrm{85}−\mathrm{1}.\mathrm{095}\right)}}{\mathrm{4}} \\ $$$${A}_{\Delta{big}} =\frac{\left(\mathrm{1}+\mathrm{20}\right)×\left(\mathrm{0}.\mathrm{85}+\mathrm{17}\right)}{\mathrm{1}×\mathrm{0}.\mathrm{85}}×\frac{\sqrt{\left(\mathrm{1}+\mathrm{0}.\mathrm{85}+\mathrm{1}.\mathrm{095}\right)\left(−\mathrm{1}+\mathrm{0}.\mathrm{85}+\mathrm{1}.\mathrm{095}\right)\left(\mathrm{1}−\mathrm{0}.\mathrm{85}+\mathrm{1}.\mathrm{095}\right)\left(\mathrm{1}+\mathrm{0}.\mathrm{85}−\mathrm{1}.\mathrm{095}\right)}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{178}.\mathrm{318} \\ $$$${or} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{0}.\mathrm{85}^{\mathrm{2}} −\mathrm{1}.\mathrm{095}^{\mathrm{2}} }{\mathrm{2}×\mathrm{1}×\mathrm{0}.\mathrm{85}}=\mathrm{0}.\mathrm{307926} \\ $$$$\mathrm{sin}\:\theta=\sqrt{\mathrm{1}−\mathrm{0}.\mathrm{307926}^{\mathrm{2}} }=\mathrm{0}.\mathrm{951410} \\ $$$${A}_{\Delta{big}} =\frac{\left(\mathrm{1}+\mathrm{20}\right)×\left(\mathrm{0}.\mathrm{85}+\mathrm{17}\right)×\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{178}.\mathrm{318} \\ $$
Commented by SonGoku last updated on 16/Jul/24
$$\mathrm{Thank}\:\mathrm{you} \\ $$
Commented by maths_plus last updated on 16/Jul/24
$$\mathrm{cool} \\ $$