Question

Alguém para ajudar?

Answer 1

Olá
Resolução:
Da figura baixo temos:
a)  $sen58^{0} = \frac{8}{3. \sqrt{10} } \\ \\ ==> = \frac{8. \sqrt{90} }{( \sqrt{90) ^{2} } } \\ \\ ==>= \frac{8.3 \sqrt{10} }{90} \\ \\ ==>= \frac{4 \sqrt{10} }{5}$

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b) $sen 32^{0} = \frac{5}{3 \sqrt{10} } = \frac{5. \sqrt{90} }{( \sqrt{90}) ^{2} } = \frac{5.3. \sqrt{10} }{90} = \frac{ \sqrt{10} }{6}$

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c) $tag302^{0} =tag(360^{0} -302^{0})=-tag58 ^{0} = - \frac{8}{5}$

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d) $tag122^{0} =tag(180^{0} -122^{0} )= -tag58^{0}= - \frac{8}{5}$

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                            Espero ter ajudado!!

Answer 2

$\mathrm{tg\,}58^{\circ}=\dfrac{8}{5}$


a) $\mathrm{sen\,}58^{\circ}$

$\alpha=58^{\circ}$

$\mathrm{tg\,}\alpha=\dfrac{\mathrm{sen\,}\alpha}{\cos \alpha}\\ \\ \mathrm{tg}^{2\,}\alpha=\dfrac{\mathrm{sen}^{2\,}\alpha}{\cos^{2}\alpha}\\ \\ \mathrm{tg}^{2\,}\alpha=\dfrac{\mathrm{sen}^{2\,}\alpha}{1-\mathrm{sen}^{2\,}\alpha}\\ \\ \mathrm{tg}^{2\,}\alpha\cdot \left(1-\mathrm{sen}^{2\,}\alpha \right)=\mathrm{sen}^{2\,}\alpha\\ \\ \mathrm{tg}^{2\,}\alpha-\mathrm{tg}^{2\,}\alpha\cdot \mathrm{sen}^{2\,}\alpha=\mathrm{sen}^{2\,}\alpha\\ \\ \mathrm{tg}^{2\,}\alpha=\mathrm{sen}^{2\,}\alpha+\mathrm{tg}^{2\,}\alpha\cdot \mathrm{sen}^{2\,}\alpha\\ \\ \mathrm{tg}^{2\,}\alpha=\mathrm{sen^{2\,}}\alpha\cdot \left(1+\mathrm{tg^{2\,}}\alpha \right )\\ \\ \mathrm{sen^{2}\,}\alpha=\dfrac{\mathrm{tg^{2\,}}\alpha}{1+\mathrm{tg^{2\,}}\alpha}\\ \\ \mathrm{sen^{2}\,}\alpha=\dfrac{\left(\,^{8}\!\!\!\diagup\!\!_{5} \right )^{2}}{1+\left(\,^{8}\!\!\!\diagup\!\!_{5} \right )^{2}}$


$\mathrm{sen^{2}\,}\alpha=\dfrac{\,^{64}\!\!\!\diagup\!\!_{25}}{1+\,^{64}\!\!\!\diagup\!\!_{25}}\\ \\ \mathrm{sen^{2}\,}\alpha=\dfrac{\,^{64}\!\!\!\diagup\!\!_{25}}{\,^{25}\!\!\!\diagup\!\!_{25}+\,^{64}\!\!\!\diagup\!\!_{25}}\\ \\ \mathrm{sen^{2}\,}\alpha=\dfrac{64}{25+64}\\ \\ \mathrm{sen^{2}\,}\alpha=\dfrac{64}{89}\\ \\ \mathrm{sen\,}\alpha=\pm \sqrt{\dfrac{64}{89}}\\ \\ \mathrm{sen\,}\alpha=\pm \dfrac{\sqrt{64}}{\sqrt{89}}\\ \\ \mathrm{sen\,}\alpha=\pm \dfrac{8}{\sqrt{89}}$


Como $0^{\circ}<\alpha<90^{\circ}$, então $\alpha=58^{\circ}$ é um arco do primeiro quadrante, e o seu seno é positivo. Sendo assim,

$\mathrm{sen\,}\alpha=\dfrac{8}{\sqrt{89}}\\ \\ \mathrm{sen\,}\alpha=\dfrac{8\cdot \sqrt{89}}{\sqrt{89}\cdot \sqrt{89}}\\ \\ \mathrm{sen\,}\alpha=\dfrac{8\sqrt{89}}{\left(\sqrt{89} \right )^{2}}\\ \\ \mathrm{sen\,}\alpha=\dfrac{8\sqrt{89}}{89}\Rightarrow\;\;\boxed{\mathrm{sen\,}58^{\circ}=\dfrac{8\sqrt{89}}{89}}$


b) $\mathrm{sen\,}32^{\circ}$

$\mathrm{sen\,}32^{\circ}=\cos \left(90^{\circ}-32^{\circ} \right )\\ \\ \mathrm{sen\,}32^{\circ}=\cos 58^{\circ}\\ \\ \mathrm{sen^{2}\,}32^{\circ}=\cos^{2} 58^{\circ}\\ \\ \mathrm{sen^{2}\,}32^{\circ}=1-\mathrm{sen^{2\,}}58^{\circ}\\ \\ \mathrm{sen^{2}\,}32^{\circ}=1-\left(\dfrac{8}{\sqrt{89}} \right )^{2}\\ \\ \mathrm{sen^{2}\,}32^{\circ}=1-\dfrac{64}{89}\\ \\ \mathrm{sen^{2}\,}32^{\circ}=\dfrac{89-64}{89}\\ \\ \mathrm{sen^{2}\,}32^{\circ}=\dfrac{25}{89}\\ \\ \mathrm{sen\,}32^{\circ}=\pm \sqrt{\dfrac{25}{89}}\\ \\ \mathrm{sen\,}32^{\circ}=\pm \dfrac{\sqrt{25}}{\sqrt{89}}\\ \\ \mathrm{sen\,}32^{\circ}=\pm \dfrac{5}{\sqrt{89}}$


Como $32^{\circ}$ é um arco do 1º quadrante, então o seu seno é positivo. Logo,

$\mathrm{sen\,}32^{\circ}=\dfrac{5}{\sqrt{89}}\\ \\ \mathrm{sen\,}32^{\circ}=\dfrac{5\cdot \sqrt{89}}{\sqrt{89}\cdot \sqrt{89}}\\ \\ \mathrm{sen\,}32^{\circ}=\dfrac{5\sqrt{89}}{\left(\sqrt{89} \right )^{2}}\\ \\ \boxed{\mathrm{sen\,}32^{\circ}=\dfrac{5\sqrt{89}}{89}}$


c) $\mathrm{tg\,}302^{\circ}$

$=\mathrm{tg}\left(360^{\circ}-58^{\circ} \right )\\ \\ =\mathrm{tg}\left(-58^{\circ}+2\cdot 180^{\circ} \right )\\ \\ =\mathrm{tg}\left(-58^{\circ} \right )\\ \\ =-\mathrm{tg\,}58^{\circ}\\ \\ =-\dfrac{8}{5}$


d) $\mathrm{tg\,}122^{\circ}$

$=\mathrm{tg}\left(180^{\circ}-58^{\circ} \right )\\ \\ =\mathrm{tg}\left(-58^{\circ}+1\cdot 180^{\circ} \right )\\ \\ -\mathrm{tg\,}58^{\circ}\\ \\ =-\dfrac{8}{5}$