given sin a = 5/13 and sin b = 4/5 and both a and b are acute, find cos (90 - a).
(cos a = 12/13 and cos b = 3/5)
2007-10-11 14:02:48 · 5 answers · asked by candy 3 in Science & Mathematics ➔ Mathematics
lol i found the cos.. i just added it so you wouldn't have to do as much work. :D
2007-10-11 14:11:31 · update #1
AND i'm not allowed to use a calculator.. so i need to find cos ((30+60) - a)?? :S
2007-10-11 14:12:50 · update #2
You should certainly know the identities:
cos a = sin (90° - a)
sin a = cos (90° - a)
So cos (90° - a) = sin a = 5/13.
2007-10-11 14:10:33 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
well that's not so bad, since you're given cos a as 12/13, you can use the arccos (or cos^-1 on a lot of calcs) funciton to find a, so arccos of 12/13 is 22.6 and change. Now take the cos of 90 - 22.6 and you have your answer;
0.385 (rounded to 3 dp) or for as a fraction; 5/13
2007-10-11 21:10:18 · answer #2 · answered by Philip H 2 · 0⤊ 2⤋
cos (90 - a) = sin(a) = 5/13
Is this a trick question? The answer is given.
cos (90-a) = sin(a) is an identity you should remember.
2007-10-11 21:15:52 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
Cos(90-A)=SinA. This law is true for all acute angles.
2007-10-11 21:11:04 · answer #4 · answered by Anonymous · 0⤊ 0⤋
oh the days of trig in college.
2007-10-11 21:05:39 · answer #5 · answered by BLASTING ON FOOLS 2 · 0⤊ 3⤋