Question

Given sin(x°) = .8, why is the Wolf giving 57.296 approx as the right triangle value for x when...

Answer 1

... the answer should be 53.130 degrees approx ?

The Wolf's calculation here: http://www.wolframalpha.com/input/?i=sin%28+x+degrees%29+%3D+.8

Answer 2

impossible

Answer 3

not quite sure what your issue is:

If you hover over the point on the graph in your link where sin x = .8 it gives a value of 53.13 deg

(note - the angle oyu give is 'almost' the complement of 53.13)

Answer 4

check this
http://www.wolframalpha.com/input/?i=sin%28+x+degrees%29+%3D+.8%2C0%3Cx%3C180

Answer 5

http://www.wolframalpha.com/input/?i=convert+arcsin%28.8%29+to+degrees

Answer 6

I understand and appreciate the postings of how I should have entered the query.

What I don't follow is what question the answer of the 57ish degrees ( Reminds me of 1 radian.)

In short, how is the "Wolf" interpreting my question as originally entered?

http://www.wolframalpha.com/input/?i=sin%28+x+degrees%29+%3D+.8

Answer 7

its not only 57.26

its approx 57.26(6.28n+2.21)

and 57.26(6.28n+0.92)

for if u substitute n=1
\(\large\tt \begin{align} \color{black}{57.26(6.2823\cdot n+0.927)\hspace{.33em}\\~\\
=57.26(6.2823+0.927)\hspace{.33em}\\~\\
=412
=412.272-360
=52.272
\approx 53
}\end{align}\)

Answer 8

If you click on the "approximate forms" in the answer it displays two possible sets of solutions:\[x\approx57.296(6.2832n+2.2143)\]and\[x\approx57.296(6.2832n+0.9273)\]If you look at the second set of solutions and use \(n=0\) you will get \(x\approx5.13^{\circ}\)

Answer 9

Sorry - typo - I meant you will get \(x\approx53.13^{\circ}\)