Kaylah666 Kaylah666
  • 15-11-2017
  • Mathematics
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Find the value of sec θ for the point (7,6)

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jdoe0001 jdoe0001
  • 15-11-2017
[tex]\bf (\stackrel{a}{7}~~,~~\stackrel{b}{6})\impliedby \textit{let's find the hypotenuse first} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{7^2+6^2}\implies c=\sqrt{49+36}\implies \boxed{c=\sqrt{85}} \\\\\\ sec(\theta)=\cfrac{hypotenuse}{adjacent}\qquad \qquad sec(\theta )=\cfrac{\sqrt{85}}{7}[/tex]
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