Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\cos \theta = \frac { 8 } { 17 } , \frac { 3 \pi } { 2 }

Question

Let θ be an angle in standard position. Name the quadrant in which the angle θ lies.
-$\cos \theta = \frac { 8 } { 17 } , \frac { 3 \pi } { 2 } < \theta < 2 \pi \quad$ Find $\cot \theta$.
A) $- \frac { 8 } { 15 }$
B) $- \frac { 8 } { 3 }$
C) $- \frac { 15 } { 8 }$
D) $\frac { 17 } { 8 }$

Quizplus · Accepted Answer

Given $\cos \theta = \frac{8}{17}$ and $\frac{3\pi}{2} < \theta < 2\pi$, $\theta$ lies in the fourth quadrant where cosine is positive and sine is negative. To find $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we first find $\sin \theta$ using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, which gives $\sin \theta = -\frac{15}{17}$ (negative because $\theta$ is in the fourth quadrant). Thus, $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{8/17}{-15/17} = -\frac{8}{15}$.

Let θ Be an Angle in Standard Position cos⁡θ=817,3π2<θ<2π\cos \theta = \frac { 8 } { 17 } , \frac { 3 \pi } { 2 } < \theta < 2 \pi \quadcosθ=178​,23π​<θ<2π

Let θ Be an Angle in Standard Position $\cos \theta = \frac { 8 } { 17 } , \frac { 3 \pi } { 2 } < \theta < 2 \pi \quad$