Question

Can someone help me in this trig question, please? thanks

Answer 1

$\bf \textit{the position of the rider is clearly }20cos\left( \frac{5\pi }{12} \right)~~,~~20sin\left( \frac{5\pi }{12} \right)\\\\ -------------------------------\\\\ \cfrac{5}{12}\implies \cfrac{2+3}{12}\implies \cfrac{2}{12}+\cfrac{3}{12}\implies \cfrac{1}{6}+\cfrac{1}{4} \\\\\\ \textit{therefore then }\qquad \cfrac{5\pi }{12}\implies \cfrac{1\pi }{6}+\cfrac{1\pi }{4}\implies \cfrac{\pi }{6}+\cfrac{\pi }{4}\\\\ -------------------------------$

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\ -------------------------------\\\\ cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=cos\left( \frac{\pi }{6}\right)cos\left(\frac{\pi }{4} \right)-sin\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)$

$\bf cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}-\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{6}}{4}-\cfrac{\sqrt{2}}{4}\implies \boxed{\cfrac{\sqrt{6}-\sqrt{2}}{4}} \\\\\\ sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=sin\left( \frac{\pi }{6}\right)cos\left( \frac{\pi }{4} \right)+cos\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)$

$\bf sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}+\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{2}}{4}+\cfrac{\sqrt{6}}{4}\implies \boxed{\cfrac{\sqrt{2}+\sqrt{6}}{4}}\\\\ -------------------------------\\\\ 20\left( \cfrac{\sqrt{6}-\sqrt{2}}{4} \right)\implies 5(-\sqrt{2}+\sqrt{6}) \\\\\\ 20\left( \cfrac{\sqrt{2}+\sqrt{6}}{4} \right)\implies 5(\sqrt{2}+\sqrt{6})$