Question

6,8 dau si coroana....​

Answer 1

Răspuns:

$6) sin^2x+cos^2x=1$

$sinx=\frac{cateta// opusa}{ipotenuza}$

$cosx=\frac{cateta// alaturata}{ipotenuza}$

$sinx=\frac{AB}{AC}, cosx=\frac{BC}{AC}$

$sin^2x+cos^2x=\frac{AB^2}{AC^2}+\frac{BC^2}{AC^2}<=>\frac{AB^2+BC^2}{AC^2} = \frac{AC^2}{AC^2}=1$

Explicatie: $AB^2+BC^2=AC^2$ - din teorema lui Pitagora

$tgx=\frac{cateta//opusa}{cateta//alaturata} \\ctg=\frac{cateta//alaturata}{cateta//opusa}$

$tgx=\frac{AB}{BC}, ctgx=\frac{BC}{AB}$

$tgx*ctgx=\frac{AB}{BC}*\frac{BC}{AB}=1$

$\frac{sinx}{cosx} =tgx <=>\frac{\frac{AB}{AC} }{\frac{BC}{AC} }=\frac{AB}{BC}=tgx$ "Adevarat"

$\frac{cosx}{sinx}=ctgx<=>\frac{\frac{BC}{AC} }{\frac{AB}{AC} }=\frac{BC}{AB}=ctgx$ "Adevarat"

$7) sinx=\frac{1}{4} = \frac{AB}{AC}=>AB=1k, AC=4k$

$BC^2=AC^2-AB^2<=>(4k)^2-(k)^2 = 16k^2-k^2=>BC^2=15k^2=>\\=>BC=k\sqrt{15}$

$cosx=\frac{BC}{AC}=\frac{k\sqrt{15} }{4k} =\frac{\sqrt{15} }{4}$

$tgx=\frac{AB}{BC}=\frac{k}{k\sqrt{15} } =\frac{\sqrt{15} }{15}$

$ctgx=\frac{BC}{AB}=\frac{k\sqrt{15} }{k}=\sqrt{15}$

Unghiul x = m(<BCA), am lucrat pe triunghiul din imagine: