Question

5 si 6
Dau coroana si 28 de puncte.
Materie clasa a 10-a.
Multumesc anticipat.​

Answer 1

Răspuns:

Explicație pas cu pas:

(m-2)²+(m+2)²=4²

2m²+8=16

2m²=8

m²=4

m∈{-2;2}

ctgx=6...cateta alaturata =6k

cateta opusa =k

⇒ipotenuza=√37k²=k√37

sinx= cateta opusa/ipotenuza= k/k√37=1/√37

sin²x=1/37

Answer 2

$\it 5)\ \ AB^2=(m-2)^2+(-2-m)^2=(m-2)^2+(m+2)^2=\\ \\ =m^2-4m+4+m^2+4m+4=2(m^2+4)\ \ \ \ \ (1)\\ \\ \\ AB=4 \RightarrowAB^2=16\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow 2(m^2+4)=16|_{:2}\Rightarrow m^2+4=8|_{-4} \Rightarrow m^2=4 \Rightarrow m=\pm2$

$\it 6)\ \ ctgx=6 \Rightarrow \dfrac{cosx}{sinx}=6 \Rightarrow cosx=6sinx \Rightarrow cos^2x=36sin^2x|_{+sin^2x} \Rightarrow \\ \\ \Rightarrow cos^2x+sin^2x=37sin^2x \Rightarrow 1=37sin^2x \Rightarrow sin^2x=\dfrac{1}{37}$