## anonymous one year ago Given the following triangle, solve for x.

1. jdoe0001

well $$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}$$|dw:1438558588866:dw| which of the SOH CAH TOA identities, include only the angle the adjacent side the hypotenuse ?

2. anonymous

I'm so confused, i'm sorry

3. jdoe0001

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?

4. anonymous

yes, I have.

5. jdoe0001

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

6. jdoe0001

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

7. jdoe0001

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

8. anonymous

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

9. jdoe0001

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