Question

How do you find the angle `alpha` such that the angle lies in quadrant II and `cotalpha=-0.3899`?

Answer 1

`110.30^o`

Explanation:

Tangent. and so, cotangent is negative in Q2. and Q4. For a

negative cotangent, the angle in Q4 [-pi/2, 0] is called the principal

value.

The Q2 value can be selected from the general value.

The reciprocal `tan alpha = -1/0.3899=-2.565`, nearly.

The principal `alpha=-68.70^o`,. from a calculator.

The general value is `npi + alpha=`(180n- -68.70)^o#, n =0. +-1, +-2, +-3..

The minimum Q2-value is obtained for n = 1 as `110.30^o`