## anonymous one year ago Exact measures

1. anonymous

2. campbell_st

so do you know able the exact value ratios in trigonometry...?

3. anonymous

yes the sin, cos and tan

4. anonymous

I get the proportions mixed up!

5. anonymous

for a i got 5sin30

6. Compassionate

@campbell_st , SOHCAHTOA isn't really a ratio. SOH = Sine OVER Hypotenuse CAH = Cosine over Hypotenuse TOA = Tanget over Adjacent

7. campbell_st

ok... so that's the easiest method... use the scaling factor of similar triangles... well here is the basic triangle |dw:1443990164897:dw| apply the scaling factor to the other sides to find the exact measures.

8. ybarrap

You can use law of sines: $$\cfrac{a}{\sin C}=\cfrac{b}{\sin B}\\ \cfrac{5}{\sin 30^\circ }=\cfrac{b}{\sin 60^\circ} \\$$ Solve for b Make sense?

9. campbell_st

so remember $2.5 = \frac{5}{2}$ so multiply by the improper fraction rather than the decimal and then you don't have to worry about trig ratios

10. Mertsj

|dw:1443990313447:dw|

11. ybarrap

Use $$\sin 60^\circ =\cfrac{\sqrt 3}{2}\\ \sin 30^\circ =\cfrac{ 1}{2}\\$$

12. campbell_st

this has become to confusing for me... I'll let others answer the question rather than bombard you with options

13. ybarrap

*correction $$\cfrac{a}{\sin \color{red} A}=\cfrac{b}{\sin B}\\ \cfrac{5}{\sin 30^\circ }=\cfrac{b}{\sin 60^\circ} \\$$

14. anonymous

ybarrap, how would I solve based off of the proportion.. it's a little confusing for me

15. Mertsj

I see from this question and the previous question you posted that you are studying the special right triangles. You are supposed to learn the relationships between the sides and angles of these special triangles so that you never have to calculate them again.

16. Compassionate

Lets say you have a triangle with Hypotenuse x, adjacent y, and opposite z. say I'm given the hypotenuse x's degree of 32 degrees, and I'm given an adjacent angle 10. Now all I can solve for is either the hypotenuse or opposite angle. So, if my test question said: find the hypotenuse I would say: Tan(32) = Adjacent/x Or: Tan(32) = 10/x

17. Mertsj

Here's the deal: In the 45-45-90 isosceles triangle, the legs are equal and the hypotenuse is the leg multiplied by sqrt(2) In the 30-60-90 triangle, the leg opposite the 30 degree angle is 1/2 the hypotenuse and the other leg is the short leg multiplied by sqrt(3)

18. Mertsj

You should memorize those two facts and practice enough that you can apply them automatically without using trig ratios or the law of sines or any of the rest of it.

19. Compassionate

@Mertsj , @campbell_st , @ybarrap -- I anonymously elect @Mertsj to help this student -- as we are confusing her with our differences in methods.

20. campbell_st

I bailed early because of the flood of information

21. anonymous

Mertsj, it's more easier for me when I actually set up proportions

22. Mertsj

Do not take the easy way out, sweetheart. Trust the voice of experience. If you memorize these two special triangles, it will be a great benefit to you in the future.

23. anonymous

ok :) now that I wrote out the facts, how would that help me? I'm just so used to writing out proportions

24. Nnesha

according to the previous question: you should include sin cos and tan ratios in your answer

25. anonymous

Nnesha, I'm not sure if i'm setting up the proportion up for a correctly. I wrote a=sin30

26. anonymous

a=5sin30*

27. phi

It might help to look at an intro to trig https://www.khanacademy.org/math/trigonometry/basic-trigonometry/basic_trig_ratios/v/basic-trigonometry that will help you figure out the ratios

28. phi

but yes sin 30 = a/c and c is 5 , so sin 30 = a/5 multiply both sides by 5 $5\sin 30 =5\cdot \frac{a}{5}\\ 5\sin 30 =a$

29. phi

people memorize sin,cos, tan for the angles 0, 30, 45, 60, and 90 that is, you (should) know sin 30 = 1/2 so a= 5* sin 30 = 5* 1/2 = 5/2

30. phi

That is using trig. For "special triangles" 30-60-90 and 45-45-90 people memorize some "short cuts"

31. phi

Are you studying trig or "special triangles" ?

32. anonymous

ok thank you :) for the length of b I got 5sqrt(3)/2 not sure if it's correct

33. anonymous

I'm studying trig

34. phi

yes, b is $$5\sqrt3/2$$ cos 30 = b/c sin 60 = b/c either way, with c=5, you will get b= 5 sqr(3)/2

35. anonymous

got it thank you phi!

36. phi

if you have time, get some popcorn and watch Khan's videos

37. anonymous

thanks for the advice :D definitely will do!