Question

Find the value of cos θ for the angle (4, -√33)

Answer 1

from the coordinates this angle is in the 4th quadrant

Hypotenuse = sqrt (4^2 + 33) = sqrt 49 = 7

cos θ = 4/7

Answer 2

`\bf (\stackrel{a}{4}~,~\stackrel{b}{-√(33)})\impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=√(a^2+b^2)\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{4^2+(-√(33))^2}\implies c=√(16+33)\implies c=√(49) \\\\\\ cos(\theta )=\cfrac{\stackrel{adjacent}{4}}{\stackrel{hypotenuse}{√(49)}}`

and now, let's rationalize the denominator,


`\bf \cfrac{4}{√(49)}\cdot \cfrac{√(49)}{√(49)}\implies \cfrac{4√(49)}{(√(49))^2}\implies \cfrac{4√(49)}{49}`