[tex]\alpha\in(90^o\,,\ 180^o)\quad\implies\quad \cos\alpha<0[/tex]
Korzystamy z tożsamości trygonometrycznej:
[tex]\sin^2\alpha+\cos^2\alpha=1[/tex]
a)
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\(\frac35)^2+\cos^2\alpha=1\\\\\frac9{25}+\cos^2\alpha=1\\\\\boxed{\cos^2\alpha=\frac{16}{25}}\quad\wedge\quad\cos\alpha<0\\\\\boxed{\cos\alpha=-\frac45}[/tex]
b)
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\(\frac23)^2+\cos^2\alpha=1\\\\\frac4{9}+\cos^2\alpha=1\\\\\boxed{\cos^2\alpha=\frac{5}{9}}\quad\wedge\quad\cos\alpha<0\\\\\boxed{\cos\alpha=-\frac{\sqrt5}3}[/tex]
c)
[tex]\sin^2\alpha+\cos^2\alpha=1\\\\(\frac{\sqrt5}5)^2+\cos^2\alpha=1\\\\\frac5{25}+\cos^2\alpha=1\\\\\boxed{\cos^2\alpha=\frac{20}{25}}\quad\wedge\quad\cos\alpha<0\\\\\boxed{\cos\alpha=-\frac{2\sqrt5}5}[/tex]
