well obviously, we didn't learn anything new this week...so here's are some of the trig functions we learned so far...
pythagorean theorem: c^2 = a^2 + b^2 (use to find the hypotenuse of a right triangle)
area of a right triangle = 1/2 bh
SOHCAHTOA (used to solve right triangles):
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
area of a non right triangle = 1/2 (leg)(leg)SIN(angle b/w)
Example: non-right triangle: HIJ (left to right)H = 65 degrees, j = 2, i = 6. Find the area.
A = (1/2)(2)(6)sin(65)
A = 5.438
area of an isoceles triangle = bh
Law of Sines(used to non-right triangles):
Sin 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
what i still don't understand is how you solve a whole trigangle using law of cosines..if someone could help me with that that would be great...!!!
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Very good blog Amy.
ReplyDeleteyou must take solving an entire triangle by using law of cosines step by step. because youre basically rotating around the triangle. as you solve for the first plugged in information you'll take what you solved and in combination with something you know before you"ll solve for another part and so on and so on
ReplyDeletei would give an example but i dont know how to put a picture of the triangle in here so
if my explanation doesnt clear anything up
ask in class and i'll be happy to show you the full break down.