Yeah so this week sucked! I did not have any time to do anything fun. The highlight of my week was going to Alicia's huge house and spending two full days to work on our bridge, but we finished it =]. The worst part of the week was the soccer tournament. Not only did I get in a fight with two guys on one team, but I could not walk after the first two games...and we played four. Well back to math.
So right now the only thing I really remember is the area of a non right triangle.
Formula:1/2(leg)(leg)sin(angle b/w)
So if you had 1/2(3)(6)sin(52) your answer would be: 7.100
Now for what I don't remember. I could really use some help remembering is the Law of Sines, and the Law of Cosines. Happy Thanksgiving...well late Thanksgiving =]
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For law of sines, the formula is sinA/a=sinB/b=sinC/c where the big letters are the angles and the little letters are the sides in triangle ABC. lets say ur looking for side c. You would do sinB/b=sinC/c and then cross multiply to where it would look like: csinB=bsinC. then you divide by sinB and the solve using ur calculator. Remember that law of sines and cosines will only be used for non right triangles and law of sines needs to have a angle/side pair. i dont remember law of cosines well but at least i helped with the law of sines
ReplyDeleteLaw of Sines
ReplyDeletesinA/a - sinB/b = sinC/c
used when you know pairs in a non-right triangle
you are setting up proportions
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
Law of Sines(used to non-right triangles):
ReplyDeleteSin A/a = Sin B/b= Sin C/c
Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.
A = 1/2 (4) (5) Sin 30 degrees
A = 10 Sin 30 degrees which is aproximately = 5
Law of Cosines (used when you can't use Law of Sines):
(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)
Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.
7^2=6^2+5^2-2(5)(6)
cos a7^2-6^2-5^2= 2(5)(6)
cos acos a= 7^2-6^2-5^2 / -2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees