I have a right triangle ABC with hypotenuse length 1. Angle A measures 30 degrees. Find all of the six trig ratios of the angle A.
If you understand what is going on in class really well, this will take you no time at all. Remember, you do no have to post a complete answer. You can start the discussion, and add on to it later. Do not assume you have to fully answer the question in the first post. It should be a conversation.
since it's a right triangle and one angle is 30 degrees then that means it's a 30,60,90 right triangle. I used the pythagorean theorem to figure out the adjacent and opposite lenghts. adj. length is 1 and opp. length is 0 because 1^2+0^2= 1^2
ReplyDeleteand for the six ratios i got....
sinA: 0/1
cosA: 1/1
tanA: 0/1
cscA:1/0
secA:1/1
cotA:1/0
feel free to correct me, i'm pretty sure it's all wrong.. lol
sinA = √3/2
ReplyDeletecosA = 1/2
tanA = √3/3
secA = 2√3/3
cosA = 2
cotA = √3
yay!
I started to do what you did cami with a^2 + b^2 = c^2, but it didnt seem right, because we weeren't given the a and b (opposite and adjacent)
ReplyDeleteohh, i see what you did. oops
ReplyDeleteWhen I first read this problem, I thought it would be impossible to find the ratios, but then I remembered a worksheet Mr. Frank gave us in class.. The "Trigonometric Ratios in Special Right Triangles" worksheet. I referred back to that and was able to find the ratios.
ReplyDeleteBefore I started I determined the opposite and adjacent sides which I got to be, opposite=0.5 & adjacent=0.5√3 Now knowing these ratios I got the following six trig ratios for angle A:
SinA = .5/1
CosA = .5√3/1
TanA = .5/.5√3
CscA = 1/.5
SecA = 1/.5√3
CotA = .5√3/.5
Since the right triangle has an angle of 60 degrees, it is a 30-60-90 triangle. The ratio for the sides of a 30-60-90 is 1:2:√3. I know that the hypotenuse is the longest side, so the legs of the triangle should be less than 1. 1 is half of 2, so all of the sides of the triangle should be reduced by half. The lengths are: 1, 0.5, and √3/2. Using these, I got these trig ratios:
ReplyDeleteSinA= 0.5
CosA= √3/2
TanA= 1/√3
CscA= 1/0.5
SecA= 2/√3
CotA= √3
I would like it if someone would double check this for me..
Hmm, Well first I started off writing down the whole thing, I made sure the hypotenuse was 1, and the other two sides are unknown(opposite and adjacent), and A is 30 degrees. Next I do the pythagorean theorem, a^2 +b^2=C^2. There is a slight problem that we only know what C is(2) and I think it's nearly impossible to evaluate. So I noticed that the triangle is a 30-60-90 triangle which means that the ratio is 1 to 2 to (square root)3. Now the problem is that the hypotenuse is supposed to be the longest leg, but its 1, so now everything has to be less than one. I got stuck after that so I searched for some help, and just like Dezire said, you would have to reduce by half so the other sides would be 1/2 and square root3/2.
ReplyDeleteNow that I know the angle measures and all the sides, what's left is to do all 6 trig ratios, o/h, a/h, o/a (then the opposites) :
ReplyDeleteSin = .5/1 Cosec: 2 (1/.5=2)
Cosin = root3/2 Sec: 2/root3
Tan= ??? Cotan= root3 (a/o = root3/2/.5 then simplifies root3/1 which simplifies to root3.
SinA- .5/1
ReplyDeleteCosA- (Root) 3/2
TanA- .5/ (Root) 3/2
CscA= 1/0.5
SecA= 2/(Root) 3
CotA=(Root)3
It is a 30-60-90 triangle.
ReplyDeletesinA = √3/2
cosA = 1/2
tanA = √3
secA = 2√3/3
cosA = 2
cotA = √3
Pretend comment
ReplyDeleteMr. Frank and Tolabi