Question

Given the following triangle, solve for x.

Answer 1

well
\(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\)

which of the SOH CAH TOA identities, include only
the angle
the adjacent side
the hypotenuse ?

Answer 2

I'm so confused, i'm sorry

Answer 3

well... have you covered \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\) yet?

Answer 4

yes, I have.

Answer 5

well.. notice, you ARE GIVEN the hypotenuse, the adjacent side, and the angle that is just the variable "x"

so, you'd want to use the identity that contains the hypotenuse, the adjacent side, and the angle only

that way you solve for the angle, and use the GIVEN ones

Answer 6

\(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad thus\qquad cos(x)=\cfrac{3.5}{4.6}
\\ \quad \\
cos^{-1}[cos(x)]=cos^{-1}\left( \cfrac{3.5}{4.6}\right)\impliedby taking\quad cos^{-1}\textit{ on both sides}
\\ \quad \\
\measuredangle x=cos^{-1}\left( \cfrac{3.5}{4.6}\right)\)

Answer 7

keep in mind that \(\bf cos^{-1} [cos(whatever)]=whatever
\\ \quad \\
sin^{-1} [sin(whatever)]=whatever
\\ \quad \\
tan^{-1} [tan(whatever)]=whatever\)

Answer 8

I got 0.760 but that isn't any of the answer choices...

Answer 9

the choices show angles in degrees, as opposed to "radians"
so, when taking the inverse cosine, make sure your calculator is in "degree" mode, not "radian" mode