Question

How do you find the angle `alpha` such that `0<=alpha<=90` and `tanalpha=1.904`?

Answer 1

`alpha=62.2911^o=62^o17'28''`

Explanation:

It is given that `alpha` is in first quadrant as `0<=alpha<=90^o`.

As `tan60^o=sqrt3=1.732`, `alpha>60^o`.

However to find exact value, one needs to look at natural tangents table, which come along with logarithmic and trigonometric ratios.
Using them one can find that `alpha=62^o17'`.

Using calculators one gets more accurate value but in decimal terms and it gives `alpha=62.2911^o` and is equivalent to `62^o17'28''`