anonymous one year ago Find the Value of x and y

1. anonymous

2. anonymous

$Side~opposite~30~degrees~=~Short~Leg~=~SL=x$$Side~opposite~60~degrees~=~Long~leg~=~LL=y$$Hypotenuse~=~H$ $\large SL~=~\frac{1}{2}H$ $\large LL~=~SL\sqrt{3}$

3. anonymous

im kinda confused tho... how would you solve it

4. xapproachesinfinity

if x is the opposite what ratio function would you use?

5. xapproachesinfinity

what do you know about sin and cosine?

6. anonymous

To solve the short leg (x), plug in H for the first formula.

7. anonymous

We don't really need sin and cosine for this one, since we have a 30-60-90 triangle. Ez mode

8. xapproachesinfinity

dear you are using actually that fact!

9. anonymous

In the diagram $\large H=4\sqrt{2}$ So plug that into $\large SL~=~\frac{1}{2}H$

10. xapproachesinfinity

where did you get those ratios you are using otherwise

11. xapproachesinfinity

those numbers you are using came form sin and cos

12. anonymous

I'm well aware of that

13. xapproachesinfinity

just wanted to point that since you said no need for sin and cos so just for the poster to be cleared where those came from

14. anonymous

1/2 * 4 (Sqrt)2

15. anonymous

Yep. So what would we get from that?

16. anonymous

2.82

17. anonymous

Yeah, although it may be easier to leave it in the form of $$2\sqrt{2}$$ for now.

18. anonymous

alright ...

19. anonymous

Now that we know that $$SL = 2\sqrt{2}$$ we can plug it into the next formula $\large LL~=~SL\sqrt{3}$

20. anonymous

so it would be 2sqrt2 * Sqrt 3 ???

21. anonymous

Yes. $\large 2\sqrt{2 \times 3}$

22. anonymous

oh i got it now !! thank you very very much LegendarySadist !!!

23. anonymous

$\huge \color{aqua}N\color{fuchsia}o \space \color{lime}P \color{orange}r \color{blue}o \color{maroon}b \color{red}l \color{olive}e \color{purple}m \ddot\smile$