trigonometry, someone help me with this
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[tex]\displaystyle\bf\\m(\sphericalangle~R)=30^o\\\\\implies~~\frac{PQ}{QR}=\frac{1}{2}\\\\QR=2\times\frac{4\sqrt{3}}{3}\\\\ \boxed{\bf~y=\frac{8\sqrt{3}}{3}}\\\\\textbf{We use the Pythagorean theorem}\\\\PR=\sqrt{QR^2-PQ^2}=\sqrt{\left(\frac{8\sqrt{3}}{3}}\right)^2-\left(\frac{4\sqrt{3}}{3}}\right)^2}=\\\\\\=\sqrt{\frac{64\times3}{9}-\frac{16\times3}{9}}}=\sqrt{\frac{64}{3}-\frac{16}{3}}}= \sqrt{\frac{64-16}{3}}=\sqrt{\frac{48}{3}}=\sqrt{16}=4\\\\\\\boxed{\bf~x=4}[/tex]