I have a right triangle with the three sides measuring: 31m, 146m, 149m And of course one of the angles is 90 degrees. How to I figure out the other two angles?
2007-01-31 09:39:21 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
use sine and cosine rules
sin@ = opposite/hypotenuse
cos@ = opposite/adjacent
since u have all the sides, just plug some numbers into either one of the equations and to an inverse sin/cos to get @
2007-01-31 09:44:51 · answer #1 · answered by wendywei85 3 · 0⤊ 0⤋
Strictly speaking you can't -- because such a triangle is impossible in the ordinary sense.
A triangle with sides of 31, 146, and 149 would have angles of 12.008 degrees, 78.473 degrees, and 89.519 degrees. None of these is 90 degrees.
Either one of the numbers in the problem is wrong, or this is not an ordinary 2-dimensional triangle.
2007-01-31 17:50:45 · answer #2 · answered by morningfoxnorth 6 · 0⤊ 0⤋
use arcsin on a calculator. to find the one across from 31, do arcsin(146/149) to get 78.5 degrees, then subtract from 90 to get third angle, turns out to be 11.5 degrees
2007-01-31 17:45:58 · answer #3 · answered by climberguy12 7 · 0⤊ 0⤋
Use the inverse of the proper trig function.
2007-01-31 17:45:09 · answer #4 · answered by shfincter S 2 · 0⤊ 0⤋
sine cosine and tangent
2007-01-31 17:44:45 · answer #5 · answered by john a 1 · 0⤊ 0⤋