Question

oblicz wartości funkcji trygonometrycznej kąta alfa , gdzie ramię końcowe przechodzi przez punkt A

a. A(11,5)
b. A(✓11 , ✓5)

prośba o pomoc w zadaniu ​

Answer 1

Verified answer

a)

[tex]A(11,5)\\\\x=11\quad y=5\\\\r=\sqrt{x^2+y^2}=\sqrt{11^2+5^2}=\sqrt{121+25}=\sqrt{146}\\\\\sin\alpha=\frac{y}{r}=\frac{5}{\sqrt{146}}=\frac{5}{\sqrt{146}}*\frac{\sqrt{146}}{\sqrt{146}}=\frac{5\sqrt{146}}{146}\\\\\cos\alpha=\frac{x}{r}=\frac{11}{\sqrt{146}}=\frac{11}{\sqrt{146}}*\frac{\sqrt{146}}{\sqrt{146}}=\frac{11\sqrt{146}}{146}\\\\\text{tg}\alpha=\frac{y}{x}=\frac{5}{11}\\\\\text{ctg}\alpha=\frac{x}{y}=\frac{11}{5}[/tex]

b)

[tex]A(\sqrt{11},\sqrt5)\\\\x=\sqrt{11}\quad y=\sqrt5\\\\r=\sqrt{x^2+y^2}=\sqrt{(\sqrt{11})^2+(\sqrt5)^2}=\sqrt{11+5}=\sqrt{16}=4\\\\\sin\alpha=\frac{y}{r}=\frac{\sqrt5}{4}\\\\\cos\alpha=\frac{x}{r}=\frac{\sqrt{11}}{4}}\\\\\text{tg}\alpha=\frac{y}{x}=\frac{\sqrt5}{\sqrt{11}}=\frac{\sqrt5}{\sqrt{11}}*\frac{\sqrt{11}}{\sqrt{11}}=\frac{\sqrt{55}}{11}\\\\\text{ctg}\alpha=\frac{y}{x}=\frac{\sqrt{11}}{\sqrt{5}}=\frac{\sqrt{11}}{\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{55}}{5}[/tex]