Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\cos \theta = \frac { 2 } { 5 } , \quad \tan \theta

Question

Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\cos \theta = \frac { 2 } { 5 } , \quad \tan \theta < 0$
Find $\sin \theta$.
A) $- \frac { \sqrt { 21 } } { 5 }$
B) $- \frac { \sqrt { 21 } } { 2 }$
C) $- \frac { 5 } { 2 }$
D) $- \sqrt { 21 }$

Quizplus · Accepted Answer

Given $\cos \theta = \frac{2}{5}$ and $\tan \theta < 0$, the angle $\theta$ must lie in the fourth quadrant because cosine is positive and tangent is negative only in the fourth quadrant. In the fourth quadrant, sine is negative. To find $\sin \theta$, use the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$. Substituting $\cos \theta = \frac{2}{5}$ gives $\sin^2 \theta = 1 - \left(\frac{2}{5}\right)^2 = 1 - \frac{4}{25} = \frac{21}{25}$. Taking the square root gives $\sin \theta = \pm \frac{\sqrt{21}}{5}$, but since $\theta$ is in the fourth quadrant where sine is negative, $\sin \theta = -\frac{\sqrt{21}}{5}$.

Let θ Be an Angle in Standard Position cos⁡θ=25,tan⁡θ<0\cos \theta = \frac { 2 } { 5 } , \quad \tan \theta < 0cosθ=52​,tanθ<0

Let θ Be an Angle in Standard Position $\cos \theta = \frac { 2 } { 5 } , \quad \tan \theta < 0$