Question Number 159005 by Tawa11 last updated on 11/Nov/21 | ||
Answered by ajfour last updated on 12/Nov/21 | ||
$${s}=\mathrm{2}{r}\mathrm{sin}\:\alpha \\ $$$${p}={r}\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{35}/\mathrm{2}}{{r}}\:=\frac{\mathrm{7}}{\mathrm{8}};\:\:\mathrm{sin}\:\beta=\frac{\sqrt{\mathrm{15}}}{\mathrm{8}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{90}°−\alpha\right)}{\mathrm{27}}=\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha+\beta\right)}{{p}} \\ $$$$\mathrm{20sin}\:\mathrm{2}\alpha\mathrm{cos}\:\alpha=\mathrm{27sin}\:\left(\mathrm{3}\alpha+\beta\right) \\ $$$$\mathrm{10sin}\:\alpha+\mathrm{37sin}\:\mathrm{3}\alpha=\mathrm{27sin}\:\left(\mathrm{3}\alpha+\beta\right)+\mathrm{27sin}\:\mathrm{3}\alpha \\ $$$$\:\:\:=\mathrm{27cos}\:\beta\left\{\mathrm{3sin}\:\alpha−\mathrm{4sin}\:^{\mathrm{3}} \alpha\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\sqrt{\mathrm{15}}}{\mathrm{7}}\left(\mathrm{4cos}\:^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha\right)\right\} \\ $$$${say}\:\:\mathrm{tan}\:\alpha={m} \\ $$$$\mathrm{40}=\mathrm{27}×\frac{\mathrm{7}}{\mathrm{8}}\left\{\mathrm{3}\left(\mathrm{1}+{m}^{\mathrm{2}} \right)−\mathrm{4}{m}^{\mathrm{3}} \right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\sqrt{\mathrm{15}}}{\mathrm{7}}\left[\mathrm{4}−\mathrm{3}\left(\mathrm{1}+{m}^{\mathrm{2}} \right)\right]\right\} \\ $$$$...{we}\:{need}\:{a}\:{straight}\:{cubic}\: \\ $$$${formula}...{still}\:{trying}! \\ $$ | ||
Commented by Tawa11 last updated on 12/Nov/21 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$ | ||
Answered by mr W last updated on 11/Nov/21 | ||
Commented by mr W last updated on 12/Nov/21 | ||
$${s}_{\mathrm{1}} {s}_{\mathrm{2}} =\mathrm{27}×\mathrm{8} \\ $$$${s}_{\mathrm{1}} +{s}_{\mathrm{2}} ={s} \\ $$$${s}_{\mathrm{2}} +\frac{\mathrm{216}}{{s}_{\mathrm{2}} }={s} \\ $$$${s}_{\mathrm{2}} ^{\mathrm{2}} −{ss}_{\mathrm{2}} =−\mathrm{216} \\ $$$${s}_{\mathrm{2}} =\frac{{s}+\sqrt{{s}^{\mathrm{2}} −\mathrm{864}}}{\mathrm{2}} \\ $$$${s}=\mathrm{2}×\mathrm{20}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\frac{{s}}{\mathrm{40}} \\ $$$$\mathrm{27}^{\mathrm{2}} ={s}^{\mathrm{2}} +{s}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}{ss}_{\mathrm{2}} \mathrm{cos}\:\theta \\ $$$$\mathrm{27}^{\mathrm{2}} ={s}^{\mathrm{2}} +{s}_{\mathrm{2}} ^{\mathrm{2}} −{ss}_{\mathrm{2}} +{ss}_{\mathrm{2}} \left(\mathrm{2cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{27}^{\mathrm{2}} +\mathrm{216}={s}^{\mathrm{2}} +{ss}_{\mathrm{2}} \left(\mathrm{2cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}−\mathrm{1}\right) \\ $$$${s}^{\mathrm{2}} +\frac{{s}^{\mathrm{2}} +{s}\sqrt{{s}^{\mathrm{2}} −\mathrm{864}}}{\mathrm{2}}\left(\frac{{s}^{\mathrm{2}} }{\mathrm{800}}−\mathrm{1}\right)=\mathrm{945} \\ $$$$\Rightarrow{s}\approx\mathrm{29}.\mathrm{8841} \\ $$ | ||
Commented by Tawa11 last updated on 12/Nov/21 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$ | ||