[tex]\alpha = 60^o , \ b= 50 \ feet \\ \\ tan\alpha =\frac{b}{a}\\ \\tan60^o = \frac{b}{50}\\ \\\sqrt{3}=\frac{b}{50}\ \ / \cdot 50\\ \\b=50\sqrt{3} \approx 50\cdot 1.73 = 86.5 \ feet \\ \\Answer : \ Frisbee \ is \ about \ 50 \ feet \ from \ the\ base \ of \ the \ tree[/tex]
