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

`cos(B) = -(12)/(13)`

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) = (5)/(13)`

So, applying the trigonometric identity:

`\sin^(2){B} + \cos^(2){B} = 1`

`((5)/(13))^(2)) + \cos^(2){B} = 1`

`\cos^(2){B} = 1 - ((5)/(13))^(2))`

`\cos^(2){B} = 1 - (25)/(169)`

`\cos^(2){B} = (169)/(169) - (25)/(169)`

`\cos^(2){B} = (144)/(169)`

`cos(B) = \pm \sqrt{(144)/(169)}`

`cos(B) = \pm (12)/(13)`

In the second quadrant, the cosine is negative. So

`cos(B) = -(12)/(13)`