Question Number 156423 by cortano last updated on 11/Oct/21
Commented by john_santu last updated on 11/Oct/21
$${answer}\:=\:\frac{\mathrm{81}\left(\sqrt{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}}\:{sq}\:{units} \\ $$
Commented by cortano last updated on 11/Oct/21
$$\mathrm{ok}\:\mathrm{both}\:\mathrm{sirs} \\ $$
Answered by mr W last updated on 11/Oct/21
$${rectangle}\:{a}×{b}\:{with}\:{b}=\mathrm{9} \\ $$$$\mathrm{tan}\:\alpha=\frac{{a}}{{b}}=\frac{{a}}{\mathrm{9}} \\ $$$$\mathrm{tan}\:\left(\alpha+\mathrm{30}\right)=\frac{{a}+\mathrm{9}}{\mathrm{9}} \\ $$$$\frac{\frac{{a}}{\mathrm{9}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{{a}}{\mathrm{9}\sqrt{\mathrm{3}}}}=\frac{{a}+\mathrm{9}}{\mathrm{9}} \\ $$$$\frac{\mathrm{9}+\sqrt{\mathrm{3}}{a}}{\mathrm{9}\sqrt{\mathrm{3}}−{a}}=\frac{{a}+\mathrm{9}}{\mathrm{9}} \\ $$$${a}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{81}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\mathrm{0} \\ $$$${a}=\frac{\mathrm{9}\left(\sqrt{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{ab}=\frac{\mathrm{81}\left(\sqrt{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}−\mathrm{1}\right)}{\mathrm{2}} \\ $$
Commented by mr W last updated on 11/Oct/21
$${no}! \\ $$$$\mathrm{tan}\:\alpha=\frac{{a}}{{b}}=\frac{{a}}{\mathrm{9}} \\ $$$$\mathrm{tan}\:\left(\alpha+\mathrm{30}\right)=\frac{{a}+\mathrm{9}}{{b}}=\frac{{a}+\mathrm{9}}{\mathrm{9}} \\ $$
Commented by cortano last updated on 11/Oct/21
$$\mathrm{typo}\:\mathrm{sir}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{tan}\:\left(\alpha−\mathrm{30}°\right) \\ $$
Commented by cortano last updated on 11/Oct/21
$$\mathrm{owh}\:\mathrm{but}\:\mathrm{the}\:\mathrm{answer}\:\frac{\mathrm{81}\sqrt{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{81}}{\mathrm{2}} \\ $$$$\mathrm{ok}.\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mr W last updated on 11/Oct/21
$${please}\:{recheck}\:{sir}!\:{i}\:{got} \\ $$$$\frac{\mathrm{81}\sqrt{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{81}}{\mathrm{2}} \\ $$$${not} \\ $$$$\frac{\mathrm{81}\sqrt{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{1}}}{\mathrm{2}}−\frac{\mathrm{81}}{\mathrm{2}} \\ $$
Commented by cortano last updated on 11/Oct/21
$$\mathrm{yes}.\:\mathrm{i}\:\mathrm{meant}\:\mathrm{it}.\:\mathrm{typo} \\ $$