I found a pre-calculus test recently and it's still blank (since I had to solve on another sheet of paper), and I want to do this test for fun since I have no life lol The problem is that I graduated Dec 2010 and this test is about 2 years old.
I just need some freshening up on my math.
Examples:
1.) A= 43 degrees, B= 30 degrees, b= 11
2.) A= 48 degrees, b= 15, c= 8
3.) a= 6, b= 12, c=8
I just need some freshening up on my math.
Examples:
1.) A= 43 degrees, B= 30 degrees, b= 11
2.) A= 48 degrees, b= 15, c= 8
3.) a= 6, b= 12, c=8
-
(1) 11/sin(30 degrees) = a/sin(43 degrees)
a = 11* sin(43 degrees)/sin(30 degrees)
C = (180 - 30 - 43) degrees = 107 degrees
c = 11* sin(107 degrees)/sin(30)
(2) a^2 = b^2 + c^2 - 2bc cos A
= 289 - 240 cos(48 degrees)
a = sqrt[ 289 - 240 cos(48 degrees) ]
After you have a, you can find B by
b^2 = a^2 + c^2 - 2ac cos B, or
cos B = (a^2 - 161)/16a
(3) a^2 = b^2 + c^2 - 2bc cos A
cos A = 172/192 => A = 26.4 degrees
cos B = -44/96 => B = 117.3 degrees
C = 180 - A - B
a = 11* sin(43 degrees)/sin(30 degrees)
C = (180 - 30 - 43) degrees = 107 degrees
c = 11* sin(107 degrees)/sin(30)
(2) a^2 = b^2 + c^2 - 2bc cos A
= 289 - 240 cos(48 degrees)
a = sqrt[ 289 - 240 cos(48 degrees) ]
After you have a, you can find B by
b^2 = a^2 + c^2 - 2ac cos B, or
cos B = (a^2 - 161)/16a
(3) a^2 = b^2 + c^2 - 2bc cos A
cos A = 172/192 => A = 26.4 degrees
cos B = -44/96 => B = 117.3 degrees
C = 180 - A - B
-
google law of sines
and
law of cosines
and
law of cosines