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PreCal math problems, I need someone to double check. (Will pay money!)

G

GRagland

Senior member
!!!anyone know de moivres theorem ( it deals with complex numbers)? please PM if you do and would be willing to help me for a minute. I will pay you! (paypal)!!!

OK, so that problem is done.
Heres another! I think i did it right:

1. Let vector v= (6, -2). If PQ (vector) is a representation of that vector and Q is located at (10,5), then where is P?

- I got P is at (4,7)

Is this right?

2. A 5-Mile straight of I-90 at the columbia basin has a 3% downhill grade. A semi truck weighing 20 tons is parked on the road. what is the force that that tends to make the truck roll downhill?

 
sohcahtoa

remember that, and it will work

sin= o/h
cos= a/h
tan=o/a

i believe i am right there am i not?

MIKE

isekkii pythagorean theorom = RIGHT triangles ONLY

and yeah this is from trig, not precalc
 
Originally posted by: nourdmrolNMT1
sohcahtoa

remember that, and it will work

sin= o/h
cos= a/h
tan=o/a

i believe i am right there am i not?

MIKE
Click to expand...

this is not a right angle so there is no hypotenuse. you have to use a different equation.
i just need someone to quickly double check the answer, thats all, i dont need help on how to do the problem.
 
there is ALWAYS a hypotenuse, the hypotenuse is the side that is opposite the largest angle, or has the longest length, dont know where you learned your math
 
Originally posted by: nourdmrolNMT1
there is ALWAYS a hypotenuse, the hypotenuse is the side that is opposite the largest angle, or has the longest length, dont know where you learned your math
Click to expand...

i'm scared of what they are teaching in schools these days...
 
i know how to do the problem, but thanks, i just want someone to double check it real quick, just in case im making a stupid mistake or something.
 
Originally posted by: nourdmrolNMT1
there is ALWAYS a hypotenuse, the hypotenuse is the side that is opposite the largest angle, or has the longest length, dont know where you learned your math
Click to expand...

you are wrong.
 
Originally posted by: AgentEL
Originally posted by: nourdmrolNMT1
there is ALWAYS a hypotenuse, the hypotenuse is the side that is opposite the largest angle, or has the longest length, dont know where you learned your math
Click to expand...

i'm scared of what they are teaching in schools these days...
Click to expand...

d00d, im only 16, i learned that a LONG time ago ,,,,

how am i wrong pip?
 
Originally posted by: nourdmrolNMT1
Originally posted by: AgentEL
Originally posted by: nourdmrolNMT1
there is ALWAYS a hypotenuse, the hypotenuse is the side that is opposite the largest angle, or has the longest length, dont know where you learned your math
Click to expand...

i'm scared of what they are teaching in schools these days...
Click to expand...

d00d, im only 16, i learned that a LONG time ago ,,,,

how am i wrong pip?
Click to expand...

i'm not pip, but it's just by definition of what a hypotenuse is. The side of a right triangle opposite of the right angle.
 
hypotenuse does not play a part in this question


you need to use the sine rule. a/SinA=b/SinB=c/SinC

as he wants to check his answer, i'm workin it out now, but it's like 11:20 here so i'm slow lol
 
ok I checked it and it works.

first cosine law to get c side.

then cosine law again to get beta and gamma angles.

one note you might not want to round early on.
 
Originally posted by: GRagland
I have to make a mock math test for my math class. If someone could double check the answers to some of the problems that im gonna put on it then that would be much appreciated.

- Solve the triange with the measures: A= 31 degrees, a=16 inches, b=23 inches. Find B, C, and c. (btw, a is the side opposite the angle A, etc...)

I got 132.2 degress for B, 16.8 degress for C, and 9 inches for c. Is this right?
Click to expand...

Yes, but you could also have 47.8 degrees for B, 101.2 degrees for C, and 30.5 inches for c.
 
is there a picuture with the question? i need to know if the 2 sides you are given are enclosing the angle so i can draw it, makin it easier to work out
 
Originally posted by: Krakerjak
Originally posted by: GRagland
I have to make a mock math test for my math class. If someone could double check the answers to some of the problems that im gonna put on it then that would be much appreciated.

- Solve the triange with the measures: A= 31 degrees, a=16 inches, b=23 inches. Find B, C, and c. (btw, a is the side opposite the angle A, etc...)

I got 132.2 degress for B, 16.8 degress for C, and 9 inches for c. Is this right?
Click to expand...

Yes, but you could also have 47.8 degrees for B, 101.2 degrees for C, and 30.5 inches for c.
Click to expand...

This is right becuse when u solve for c side you use a quadratic equation (c^2-34.4297+273=0) and there are two solutions for side c. one is 8.95 as the original poster stated and wanted us to check and the other is 30.5.


 
Originally posted by: acidvoodoo
is there a picuture with the question? i need to know if the 2 sides you are given are enclosing the angle so i can draw it, makin it easier to work out
Click to expand...

I think he is following the standard convention that the capital letter stands for the angle and the lowercase stands for the side opposite of it. There is an angle "A" and side "a" is the leg opposite of it. Same goes for B and C. You should be able to construct a picture out of the information already given.
 
Originally posted by: Krakerjak
Originally posted by: GRagland
I have to make a mock math test for my math class. If someone could double check the answers to some of the problems that im gonna put on it then that would be much appreciated.

- Solve the triange with the measures: A= 31 degrees, a=16 inches, b=23 inches. Find B, C, and c. (btw, a is the side opposite the angle A, etc...)

I got 132.2 degress for B, 16.8 degress for C, and 9 inches for c. Is this right?
Click to expand...

Yes, but you could also have 47.8 degrees for B, 101.2 degrees for C, and 30.5 inches for c.
Click to expand...

B cant be 47.8, because there is a picture and its a more then 90 degree angle. so i subtracted 47.8 from 180 which equals 132.2 degrees. is that alright to do?

 
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