Trig help again...

1105

I AM the random post master...bow down
May 3, 2003
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Daytona Bch, FL
Find the indicated power using DeMoivre's Theorem.

(sqrt(3)-i)^10

ans in trig form

ans in simplified standard form (A + Bi)


Find the fourth roots of 81i. Ans's in trig form
 
1105 said:
Find the indicated power using DeMoivre's Theorem.

(sqrt(3)-i)^10

ans in trig form

ans in simplified standard form (A + Bi)


Find the fourth roots of 81i. Ans's in trig form

hope im not too late... i know you have your final tomorrow / today.

Anyway... DeMoivres theroem...

(sqrt(3)-i)^10
tan(y/x) = tan(-1/sqrt(3)) = 11pi/6
sqrt(3)^2+(-1)^2=4
4*10(cos((11pi/6)*10) + isin((11pi/6)*10))

If I remember correctly that should work.

I think simplifying refines it to....
40(cos(55pi/3) + isin(55pi/3))

Then use your calc to figure cos and sin and thats your simplified.
 
How did you make it out of college RC? :lol:

Finally done with that class! I think I did good on the final. There were some I wasnt sure about, others were so easy it wasnt funny...

here's one, anyone can figure it out... Guy is flying a kyte with a 500ft string at an angle of elevation of 60*. How high is the kyte off the ground?

1. 577 feet, 2. 500 feet, 3. 430 feet? :lol: I love thoes questions
 
1105 said:
How did you make it out of college RC? :lol:

Finally done with that class! I think I did good on the final. There were some I wasnt sure about, others were so easy it wasnt funny...

here's one, anyone can figure it out... Guy is flying a kyte with a 500ft string at an angle of elevation of 60*. How high is the kyte off the ground?

1. 577 feet, 2. 500 feet, 3. 430 feet? :lol: I love thoes questions
that one's easy ... 447 feet (if he is using all 500 feet of the string)
 
BlackVert said:
that one's easy ... 447 feet (if he is using all 500 feet of the string)
even if he was using all of it... the kyte would have to be right above him to be 500feet in the air, which would be 90* of elevation, and he would have to be standing on top of a roof to have it up to 577 feet
 
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1105 said:
even if he was using all of it... the kyte would have to be right above him to be 500feet in the air, which would be 90* of elevation, and he would have to be standing on top of a roof to have it up to 577 feet
yeah, but it is not 430 feet, it is 447 feet, unless the string is curved ... ugh
 
heh... angle of elevation... would be from the guy on the ground to the kite... so the kite is at a 60 degree angle from him.

If he is using 500 ft of string... then...

sin(x/500) = 60

since opposite is sin, and the vertical distance of the kite from the ground.

to solve...

arcsin(60) = x/500
x=433ft
 
1105 said:
How did you make it out of college RC? :lol:

Finally done with that class! I think I did good on the final. There were some I wasnt sure about, others were so easy it wasnt funny...

here's one, anyone can figure it out... Guy is flying a kyte with a 500ft string at an angle of elevation of 60*. How high is the kyte off the ground?

1. 577 feet, 2. 500 feet, 3. 430 feet? :lol: I love thoes questions

that's a funny question... promotes you to think logically i guess, i probably would have done the calculation anyways, just to make sure the teacher didn't screw up

on my engineering midterm the other day, a bunch of people were asking him questions, so the dumb guys says "ok, all your questions are straightforward, any other dumb questions and you automatically get -2 points, for a good question +1" -i thought that was pretty hilarious
 
yea.. none...

I forgot the exact answers for the problem, but 2 were defently not the answer and you could tell just by looking at it... The other answer was 400 something