Exact measures

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so do you know able the exact value ratios in trigonometry...?
yes the sin, cos and tan

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I get the proportions mixed up!
for a i got 5sin30
@campbell_st , SOHCAHTOA isn't really a ratio. SOH = Sine OVER Hypotenuse CAH = Cosine over Hypotenuse TOA = Tanget over Adjacent
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.
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?
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
|dw:1443990313447:dw|
Use $$ \sin 60^\circ =\cfrac{\sqrt 3}{2}\\ \sin 30^\circ =\cfrac{ 1}{2}\\ $$
this has become to confusing for me... I'll let others answer the question rather than bombard you with options
*correction $$ \cfrac{a}{\sin \color{red} A}=\cfrac{b}{\sin B}\\ \cfrac{5}{\sin 30^\circ }=\cfrac{b}{\sin 60^\circ} \\ $$
ybarrap, how would I solve based off of the proportion.. it's a little confusing for me
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.
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
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)
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.
@Mertsj , @campbell_st , @ybarrap -- I anonymously elect @Mertsj to help this student -- as we are confusing her with our differences in methods.
I bailed early because of the flood of information
Mertsj, it's more easier for me when I actually set up proportions
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.
ok :) now that I wrote out the facts, how would that help me? I'm just so used to writing out proportions
according to the previous question: you should include sin cos and tan ratios in your answer
Nnesha, I'm not sure if i'm setting up the proportion up for a correctly. I wrote a=sin30
a=5sin30*
  • 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
  • phi
but yes sin 30 = a/c and c is 5 , so sin 30 = a/5 multiply both sides by 5 \[ 5\sin 30 =5\cdot \frac{a}{5}\\ 5\sin 30 =a \]
  • 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
  • phi
That is using trig. For "special triangles" 30-60-90 and 45-45-90 people memorize some "short cuts"
  • phi
Are you studying trig or "special triangles" ?
ok thank you :) for the length of b I got 5sqrt(3)/2 not sure if it's correct
I'm studying trig
  • 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
got it thank you phi!
  • phi
if you have time, get some popcorn and watch Khan's videos
thanks for the advice :D definitely will do!

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