Geometry help

[BlckMgk]

Its been a while since I did geometry in school but I need some help with some right triangle measurements.

I have a leg length of 2 inches and 4 inches and they intersect at a right angle (90degrees)

What I need to know is:
The angle where the hypotnous and the 4inch leg meet.
The angle where the 2inch leg and the hypotnous meet.
And the length of the hypotnous.

From trying to refresh by searching the web I found the Sine, Cosine, and Tangent formulas but forgot how to apply them.

Please help!

-Blck

[megadud]

45 you're welcome

[vorticon]

short leg angle where it meets the hypotenuse is 28.64 degrees (1/ (2/4 *tan))
the other angle is 64.36 degrees (90 - 28.64)

the length is 4. 47(4/ sine(64.36))


but dont take my word for it, i was never to great at it when i was doing every day, let alone after a couple years of not doing it at all (or actually, letting the fancy construction calculator figure it out for me)

[john9001]

my calculating machine says

hypotenuse= 4.4721

angle for 4 inch leg=26.565

angle for 2 inch leg=63.435

[BlckMgk]

you guys got the same length, but different angles, how come?

-Blck..

BTW thanks!

[kamilyun]

a ------- 4 in ------------------- b
|                                      
                               
2                          
i                  
n            
           
|      
c

Length of hyp = (4^2 + 2^2)^1/2 = 20^1/2 = 2*5^1/2... "the square root of 20" or "two times the square root of five"

soh cah toa

sin = opposite over hyp
cos = adjacent over hyp
tan = opposite over adjacent

tan b = 2/4 = 0.5; b = tan^-1 (0.5) = i don't have my nice calc with me
tan c = 4/2 = 2; c = tan^-1 of (2.0) = ditto

Edit:  sweet online scientific calculator

inverse tanget of 0.5 = 26.57 degrees, this is angle b
inverse tanget of 2.0 = 63.43 degrees, this is angle c

They add up to 90...

[Dux]

SOHCAHTOA!

Glad somebody else remembers that.

[vorticon]

Originally posted by BlckMgk
you guys got the same length, but different angles, how come?

-Blck..

BTW thanks!

i was using the windows scientific calculator, and probably pressed the buttons in the wrong order.

[SIK1]

Using pathagareom therum (2^2+4^2)= hypotonus^2.

Hypotonus= 4.47"

using the law of sines the small angle is =27*
again using the law of sines the large angle is =63*

[FastFwd]

Originally posted by john9001
my calculating machine says

hypotenuse= 4.4721

angle for 4 inch leg=26.565

angle for 2 inch leg=63.435

Agreed.

[storch]

Originally posted by john9001
my calculating machine says

hypotenuse= 4.4721

angle for 4 inch leg=26.565

angle for 2 inch leg=63.435

yup

[SIK1]

You actually don't have enough significant figures to have anything less than a whole degree. Truthfuly the length of the hypotanus should be no more than two sig figs rounding up to 4.5".

[Jackal1]

OK, while you have those  brains fired up.

A cylindrical container that is 5 1/2 " tall with a circumfrence of 6 1/2 "............
How many ounces will it hold?

[Mak333]

Originally posted by Jackal1
OK, while you have those  brains fired up.

A cylindrical container that is 5 1/2 " tall with a circumfrence of 6 1/2 "............
How many ounces will it hold?

First off circumference is 2xPixRadius = 6.5

You need to find the area which is Pi x Radius^2

6.5/(2xPi) = Radius = 1.03 inches

(pi)(1.03^2)

3.36 is the area of the base

(3.36)(5.5) = 18.5 inches squared (volume)

I am not going to go any further because I have to go to my speech 101 class, but someone can figure out the ounces.

[SIK1]

Hmmm, I get 3.33"^2 for the area of the circle, and 18.3"^3 for the volume of the cylinder.

1 fl. oz.= 1.804"^3 so if you take the volume of the cylinder 18.3"^3 divide by 1.804"^3 you will get the fluid ounces that the cylinder will hold. I get 10.1 fl oz.