Question

\tan(a) = -\frac{15}{8} \text{ and } \cos(a) < 0. \text{ find } \sin(a) \text{ and } \cos(a).

\bigcirc \sin(a) = \frac{15}{17}, \cos(a) = -\frac{8}{17}
\bigcirc \sin(a) = \frac{15}{17}, \cos(a) = \frac{8}{17}
\bigcirc \sin(a) = \frac{8}{17}, \cos(a) = -\frac{15}{17}
\bigcirc \sin(a) = -\frac{15}{17}, \cos(a) = -\frac{8}{17}

Explanation:

Determine the quadrant of angle A

Given:

$$ \tan(A) = -\frac{15}{8} < 0 \quad \text{and} \quad \cos(A) < 0 $$

Since \(\tan(A) < 0\) and \(\cos(A) < 0\), the angle \(A\) must lie in Quadrant II.
In Quadrant II, \(\sin(A) > 0\).

Calculate the hypotenuse and trigonometric ratios

Let the reference right triangle have opposite side \(y = 15\) and adjacent side \(x = -8\).
Using the Pythagorean theorem:

$$ r = \sqrt{x^2 + y^2} = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 $$

Thus:

$$ \sin(A) = \frac{y}{r} = \frac{15}{17} $$

$$ \cos(A) = \frac{x}{r} = -\frac{8}{17} $$

Answer:

• (A) \(\sin(A) = \frac{15}{17}, \cos(A) = -\frac{8}{17}\) (Correct answer)
• (B) \(\sin(A) = \frac{15}{17}, \cos(A) = \frac{8}{17}\)
• (C) \(\sin(A) = \frac{8}{17}, \cos(A) = -\frac{15}{17}\)
• (D) \(\sin(A) = -\frac{15}{17}, \cos(A) = -\frac{8}{17}\)