Question

Va rog sa ma ajutati. Cu explicatie. Ambele probleme

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

6.

$\it Not\breve am: \ \ BC = a, \ \ AC = b, \ \ AB = c, \ \ m(\hat C) = x\\ \\ Din \ \ teorema \ \ lui \ \ Pitagora\ \Rightarrow \ a^2 + c^2 = b^2\ \ \ \ \ (*)$

$\it \left\begin{aligned}sinx=\dfrac{c}{b} \Rightarrow sin^2x=\dfrac{c^2}{b^2}\\ \\ cosx=\dfrac{a}{b} \Rightarrow cos^2x=\dfrac{a^2}{b^2}\end{aligned}\right\} \Rightarrow sin^2x+cos^2x=\dfrac{c^2}{b^2}+\dfrac{a^2}{b^2}=\dfrac{c^2+a^2}{b^2}\stackrel{(*)}{=}\dfrac{b^2}{b^2}=1$

$\it tgx\cdot ctgx=\dfrac{c}{a}\cdot\dfrac{a}{c}=\dfrac{ac}{ac}=1$

$\it \dfrac{sinx}{cosx}=\dfrac{\dfrac{c}{b}}{\dfrac{a}{b}}=\dfrac{c}{b}:\dfrac{a}{b}=\dfrac{c}{\not b}\cdot\dfrac{\not b}{a}=\dfrac{c}{a}=tgx \Rightarrow \dfrac{sinx}{cosx}=tgx$

7.

$\it sinx=\dfrac{1}{4}\\ \\ sin^2x+cos^2x=1 \Rightarrow cos^2x=1-sin^2x \Rightarrow cosx=\sqrt{1-sin^2x} \Rightarrow \\ \\ \Rightarrow cosx=\sqrt{1-\Big(\dfrac{1}{4}\Big)^2}=\sqrt{1-\dfrac{1}{16}}=\sqrt{\dfrac{15}{16}}=\dfrac{\sqrt{15}}{4}$

$\it tgx=\dfrac{sinx}{cosx}=\dfrac{\dfrac{1}{4}}{\dfrac{\sqrt{15}}{4}}=\dfrac{1}{4}\cdot\dfrac{4}{\sqrt{15}}=\dfrac{^{\sqrt{15})}1}{\ \ \sqrt{15}}=\dfrac{\sqrt{15}}{15}$

$\it ctgx=\dfrac{1}{tgx}=\dfrac{1}{\dfrac{1}{\sqrt{15}}}=1\cdot\dfrac{\sqrt{15}}{1}=\sqrt{15}$