Question

Помогите пожалуйста

Answer 1

$sin\alpha = -0.6; \pi < \alpha < \frac{3\pi}2 => cos\alpha < 0\\cos\alpha = \sqrt{1-sin^2\alpha} = \sqrt{1-0.36} = \sqrt{0.64}\\cos\alpha = -0.8\\tg\alpha = \frac{-0.6}{-0.8} = 0.75\\ctg\alpha = \frac{-0.8}{-0.6} = 1.(3)$

Answer 2

I hope this helps you