## Trigonometry 7th Edition

$tan(\frac{A}{2})=\frac{3-\sqrt{5}}{2}$
Note$\;\;\;\;\;\;\;sinA=\frac{2}{3}\;\;\;;\;\;\;\;0^o\leq A\leq 90^o$ $sin^2(A)+cos^2(A)=1\;\;\;\;\;\;\;\;\;\;\;\;$ and $sin(A)=\frac{2}{3}$ $(\frac{2}{3})^2+cos^2(A)=1$ $cos^2(A)=\frac{5}{9}$ $cos(A)=\pm \frac{\sqrt{5}}{3}$ We know $0^o\leq A\leq 90^o$ $cos(A)=\frac{\sqrt{5}}{3}$ $tan(\frac{A}{2})= \frac{1+cos(A)}{sin(A)}$ $tan(\frac{A}{2})=\frac{1+\frac{\sqrt{5}}{3}}{\frac{2}{3}}=\frac{3-\sqrt{5}}{2}$ $tan(\frac{A}{2})=\frac{3-\sqrt{5}}{2}$