Question

If cos θ = .62 and if 0<θ<90 what is the approximate value of sin θ to the nearest hundredth

Answer 1

$\cos\Theta=0.62\\\\\text{Use:}\ \sin^2\alpha+\cos^2\alpha=1\\\\\sin^2\Theta+(0.62)^2=1\\\\\sin^2\Theta+0.3844=1\ \ \ |-0.3844\\\\\sin^2\Theta=0.6156\to\sin\Theta=\pm\sqrt{0.6156}\\\\\sin\Theta_1\approx-0.78\ \vee\ \sin\Theta\approx0.78\\\\\Theta\in(0^o;\ 90^o),\ therefore\ \sin\Theta \ \textgreater \ 0\to\boxed{\sin\Theta\approx0.78}$