Question

для острого угла A найдите sin a, cos a, ctg a , если tg A=√3​

Answer 1

Объяснение:

$tg\alpha =\sqrt{3} \ \ \ \ 0^0<\alpha <90^0 \ \ \ \ sin\alpha =?\ \ \ \ cos\alpha =?\ \ \ \ ctg\alpha =?\\1)\ tg^2\alpha =\frac{sin^2\alpha }{cos^2\alpha } =(\sqrt{3})^2=3. \\sin^2\alpha =3*cos^2\alpha \\sin^2\alpha +cos^2\alpha=3*cos^2\alpha +cos^2\alpha \\4*cos^2\alpha =1\ |:4\\cos^2\alpha =\frac{1}{4}\\cos\alpha =б\sqrt{\frac{1}{4} }=б\frac{1}{2} \\cos\alpha = \frac{1}{2} \ \ \ \ (0^0<\alpha <90^0).$

$sin^2\alpha +cos^2\alpha =1\\sin^2\alpha =1-cos^2\alpha =1-(\frac{1}{2} )^2=1-\frac{1}{4} =\frac{3}{4}\\sin\alpha =б\sqrt{\frac{3}{4} }=б\frac{\sqrt{3} }{2}\\sin\alpha =\frac{\sqrt{3} }{2} \ \ \ \ (0^0<\alpha <90^0).$

$ctg\alpha =\frac{1}{tg\alpha } =\frac{1}{\sqrt{3} } =\frac{\sqrt{3} }{3}.$

$2)\ tg\alpha =\sqrt{3} \ \ \ \ 0^0<\alpha <90^0\ \ \ \ \ \Rightarrow\ \ \ \ \alpha =60^0.\\sin60^0=\frac{\sqrt{3} }{2}.\\cos60^0=\frac{1}{2}.\\ctg60^0=\frac{\sqrt{3} }{3}.$