MHB Given value of cos(A) and quadrant of A, find sin(A) and tan(A)

[DeusAbscondus]

Hi folks,

$\text{ Here is the problem: }$

$\text{ Given that } cosA = -\frac{1}{4} \text{ and given that A }$

$\text{ is an angle between 90 and 180 degrees, then find: }$

a) $tanA$ and

b) $sinA$

$\text{Here is my working out: }$

$\text{1. Since A is in the second quadrant, then tanA will be negative and sinA will be}$

$\text{positive.}$

$\text{2. Given that }cosA=-\frac{1}{4}$

$\text{then } -1^2+x^{2}=4^{2}\ and \ x^{2}=15$

$\therefore x=\pm \sqrt{15}\ \text{ but the only real answer here is: }x=\sqrt{15}$

$\text{Now, since }x=\text{opp side, }$

$\text{ and } -1=\text{adjacent side}$

$\text{ then: a) }tanA=\frac{\sqrt{15}}{-1}$

$\text{and b) }sinA=\frac{\sqrt{15}}{4}$

$\text{If this is right, then my textbook is wrong, since it gives the answers: }$

a) $tanA=-\frac{1}{\sqrt{15}}$

b) $sinA=\frac{1}{\sqrt{15}}$

$\text{Who is right? }$

$\text{Thanks, as always, DeusAbscondus}$

 

[chisigma]

Re: basic trig problem

DeusAbscondus - Textbook: 1-0... congratulations!(Clapping)...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

Re: basic trig problem

I agree with you:

$$\sin(A)=\sqrt{1-\cos^2(A)}=\frac{\sqrt{15}}{4}$$

$$\tan(A)=\frac{\sin(A)}{\cos(A)}=-\sqrt{15}$$