Question Number 180858 by Mastermind last updated on 18/Nov/22
Answered by mr W last updated on 18/Nov/22
$${A}=\mathrm{2}×\frac{{a}×{a}×\mathrm{tan}\:\left[\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\alpha\right)\right]}{\mathrm{2}} \\ $$$$\:\:\:={a}^{\mathrm{2}} \mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right) \\ $$
Commented by Mastermind last updated on 18/Nov/22
$$\mathrm{With}\:\mathrm{full}\:\mathrm{details}\:\mathrm{explanation}\:\mathrm{boss} \\ $$
Commented by mr W last updated on 18/Nov/22
Commented by mr W last updated on 18/Nov/22
$$\mathrm{2}\beta+\alpha=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}} \\ $$$${b}={a}\:\mathrm{tan}\:\alpha={a}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\frac{{A}}{\mathrm{2}}=\frac{{ab}}{\mathrm{2}}=\frac{{a}×{a}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{A}={a}^{\mathrm{2}} \mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right) \\ $$