+845 Help me with this maths question plz also explain why thanks
https://snag.gy/NiUh6u.jpg
+9466 
m∠RAB = 160° - 90° = 70°
m∠ABR = 360° - 90° - 220° = 50°
m∠ARB = 180° - 70° - 50° = 60°
m∠TAR = 180° - 160° = 20°
Now we can use the law of sines to find the length of AR .
\(\frac{AR}{\sin 50°}\,=\,\frac{12}{\sin60°} \\~\\ AR\,=\,\frac{12\sin50°}{\sin60°}\)
Now we can use the law of cosines to find the length of TR.
\(TR^2\,=\,AT^2+AR^2-2(AT)(AR)\cos20° \\~\\ TR^2\,=\,5^2+(\frac{12\sin50°}{\sin60°})^2-2(5)(\frac{12\sin50°}{\sin60°})\cos20° \\~\\ TR^2\,=\,25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°} \)
Take the positive sqrt of both sides.
\(TR\,=\,\sqrt{25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°}} \\~\\ TR\,\approx\,6.158\quad\text{km}\)
+9466
m∠RAB = 160° - 90° = 70°
m∠ABR = 360° - 90° - 220° = 50°
m∠ARB = 180° - 70° - 50° = 60°
m∠TAR = 180° - 160° = 20°
Now we can use the law of sines to find the length of AR .
\(\frac{AR}{\sin 50°}\,=\,\frac{12}{\sin60°} \\~\\ AR\,=\,\frac{12\sin50°}{\sin60°}\)
Now we can use the law of cosines to find the length of TR.
\(TR^2\,=\,AT^2+AR^2-2(AT)(AR)\cos20° \\~\\ TR^2\,=\,5^2+(\frac{12\sin50°}{\sin60°})^2-2(5)(\frac{12\sin50°}{\sin60°})\cos20° \\~\\ TR^2\,=\,25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°} \)
Take the positive sqrt of both sides.
\(TR\,=\,\sqrt{25+\frac{144\sin^250°}{\sin^260°}-\frac{120\sin50°\cos20°}{\sin60°}} \\~\\ TR\,\approx\,6.158\quad\text{km}\)
