Question

When drawn in standard position, an angle B has a terminal ray that lies in the second quadrant and sin(B)=5/13. Find the value of cos(B), (show your answer as a fraction). Please show all your work

Please a help here.
Thank you

Answer 1

Answer:

$\cos{B} = -\frac{12}{13}$

Step-by-step explanation:

For any angle B, we have the following trigonometric identity:

$\sin^{2}{B} + \cos^{2}{B} = 1$

In this problem, we have that:

$\sin{B} = \frac{5}{13}$

So, applying the trigonometric identity:

$\sin^{2}{B} + \cos^{2}{B} = 1$

$(\frac{5}{13})^{2}) + \cos^{2}{B} = 1$

$\cos^{2}{B} = 1 - (\frac{5}{13})^{2})$

$\cos^{2}{B} = 1 - \frac{25}{169}$

$\cos^{2}{B} = \frac{169}{169} - \frac{25}{169}$

$\cos^{2}{B} = \frac{144}{169}$

$\cos{B} = \pm \sqrt{\frac{144}{169}}$

$\cos{B} = \pm \frac{12}{13}$

In the second quadrant, the cosine is negative. So

$\cos{B} = -\frac{12}{13}$