cara mencari tan 52,5 (pake jalan cara nya ya) mohon bantuanyaa,untuk besok

[tex]
\text{Misalkan }\tan{52.5\degree}=x\\
\text{Selanjutnya, perhatikan bahwa}\\
\tan{105\degree}
=\tan{(2\cdot52.5\degree)}\\
=\frac{2\tan{52.5\degree}}{1-\tan^2{52.5\degree}}\\
=\frac{2x}{1-x^2}\\
\text{Sekarang, perhatikan bahwa}\\
\tan{105\degree}\\
=\tan{(90\degree+15\degree)}\\
=-\cot{15\degree}\\
=-\cot{(45\degree-30\degree)}\\
=-\frac{1+\tan{45\degree}\cdot\tan{30\degree}}{\tan{45\degree}-\tan{30\degree}}\\
=-\frac{1+1\cdot\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\\
=-\frac{\sqrt{3}+1}{\sqrt{3}-1}\\
\text{Dengan demikian, diperoleh kesamaan}\\
-\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{2x}{1-x^2}\\
\Leftrightarrow-\sqrt{3}-1+(\sqrt{3}+1)x^2=2(\sqrt{3}-1)x\\
\Leftrightarrow(\sqrt{3}+1)x^2-2(\sqrt{3}-1)x-\sqrt{3}-1=0\\
\text{well, dengan menggunakan rumus kuadrat (dan sedikit kesabaran) akan diperoleh}\\
x = \tan{52.5\degree}=\frac{-1+\sqrt{3}+\sqrt{20+6\sqrt{3}}}{2(1+\sqrt{3})}\\
[/tex]