What are the ratios of the sides of a right triangle with theta, an acute angle, as one of its angles if cos theta = 25? Please explain your answer. Thanks!
2007-09-04 07:49:39 · 3 answers · asked by sillyboys_trucksare4girls 2 in Science & Mathematics ➔ Mathematics
Consider a right angle triangle ABC where AC is hypotenuse and angle ABC is a right angle.
Hyoptenuse , AC = 5
BC = 2
cos C = 2 / 5
5 ² = 2 ² + AB ²
AB ² = 21
AB = √ 21
sin C = √ 21 / 5
tan C = √ 21 / 2
sin A = 2 / 5
cos A = √ 21 / 5
tan A = 2 / √ 21
2007-09-08 06:56:02 · answer #1 · answered by Como 7 · 0⤊ 0⤋
A cosine has to be in the range [-1, 1].
There is no possible triangle with cos(theta) = 25.
2007-09-04 14:58:07 · answer #2 · answered by Anonymous · 0⤊ 0⤋
cosine is equal to the ratio of the adjacent side over the hypotenuse (remember SOHCAHTOA) From that you can say that the ratio of A/H = 25. You can then use the formula a2+b2=c2, using a = 25 and c = 1. You can then find the rest of your ratios..
You may want to check your numbers, because this will leave you with an imaginary value for b. But you can use this same method to solve any "ratio" problem
2007-09-04 15:01:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋