## anonymous one year ago Solve for x. Round your answer to 2 decimal places. (picture below) ** medal & fan

1. anonymous

2. Jamierox4ev3r

do you know what the law of sines is?

3. anonymous

um, i dont think i have covered that yet.

4. Jamierox4ev3r

Does this look familiar to you at all? $$\large\frac{a}{Sin A}=\large\frac{b}{Sin B}=\large\frac{c}{Sin C}$$

5. anonymous

no, i definitely have not covered that yet :-(

6. Jamierox4ev3r

fair enough. I'm not sure how I would solve a problem like this without that, sorry

7. jdoe0001

|dw:1440460690118:dw| recall you SOH CAH TOA $$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent}$$ which identity uses the angle adjacent side, and hypotenuse only?

8. anonymous

cosine

9. jdoe0001

yeap thus... one sec

10. anonymous

i will wait for you :-)

11. jdoe0001

$$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies hypotenuse=\cfrac{adjacent}{cos(\theta)}\qquad thus \\ \quad \\ cos(58^o)=\cfrac{17}{x}\implies x=\cfrac{17}{cos(48^o)}$$

12. jdoe0001

hmmm my 58 turned into a 48 for whatever reason =) $$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies hypotenuse=\cfrac{adjacent}{cos(\theta)}\qquad thus \\ \quad \\ cos(58^o)=\cfrac{17}{x}\implies x=\cfrac{17}{cos(58^o)}$$

13. jdoe0001

notice, you're using degrees, thus, make sure your calculator is in Degree mode when getting the cosine

14. anonymous

x = 17 / 0.53 ?

15. jdoe0001

yeap

16. anonymous

i got it!! thank you so much!! that was really simple!!