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

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Originally posted by: GRagland
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?

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

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?

does it say the drawing is accurate?

and how can you do that, because you are only told one angle in the first place
 
Using law of cosines you end up with a quadratic equation. Then you employ the quadratic formula to solve for side c.
 
Originally posted by: acidvoodoo
i got

B=47
C=102
c=28

i rounded though

C cant equal 102 because there is a picture of the triangle and it looks less 90 degrees. and B is more then 90 degrees (its a big angle on the picture).

 
it'd be far easier if we could see the picture so we know what encloses what etc


and the quadratic formula is not needed in this case, this is very simple stuff
 
no matter what here are some hints to those of you trying..

angle B will be the largest angle, due to side B being the largest side.

once you find angle B you can find C by subtraction

then use quadratic to find side c
 
> it'd be far easier if we could see the picture so we know what encloses what etc

pic will not be necessary.

just follow the rules.

http://www.ping.be/~ping1339/gonio.htm#Basic-formulas


In any triangle ABC we have

a / sin(A) = b / sin(B) = c / sin(C)

In any triangle ABC we have

a2 = b2 + c2 - 2 b c cos(A)

b2 = c2 + a2 - 2 c a cos(B)

c2 = a2 + b2 - 2 a b cos(C)



or use the one acidvoodoo post...
 
Yeah i followed those rules and got my answer, but I got 47.8 degreed for B, and since i know B is the largest angle i subtracted 47.8 from 180 and got 132.2 degrees for B. I remember hearing that if you get too small an angle measure then subtract it from 180.... so is this right?
 
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?
 
use a2 = b2 + c2 - 2 b c cos(A)

you will get a quadratic c2 - 39.43c + 273 = 0

solutions are ~9 and ~30.5

Solve for rest of triangle

For the next one i just used v = P - Q ....... or v = Q - P
one will give (16,3) and (4,7)
 
ok thanks!
next question:

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?

 
dude, enough question.

seems like a physic question more than maths question.

how old are u and what kind of crap mock exam don't provide answer ?

F = m x g x sin degree

F = force
m = mass of truck
g = gravity
degree = angle of 3% hill

GRagland <- @sshat can't shut his mouth up..

google is your friend.
 
Originally posted by: jimmyhaha
dude, enough question.

seems like a physic question more than maths question.

F = m x g x sin degree

F = force
m = mass of truck
g = gravity
degree = angle of 3% hill

okay thanks, following that i got 1199.4578 lbs force pushing the truck down the hill. someone want to double check 🙂
 
anyone know de moivres theorem (deals with complex numbers)? please PM if you do and would be willing to help me for a minute. I will pay you! (paypal).
 
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