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

Aug 12, 2016

${110.30}^{o}$

#### 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 $\tan \alpha = - \frac{1}{0.3899} = - 2.565$, nearly.

The principal $\alpha = - {68.70}^{o}$,. from a calculator.

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

The minimum Q2-value is obtained for n = 1 as ${110.30}^{o}$