The terminal side of Ø is in quadrant II and sin ø = 5/13 What is cos ø?

A. 5/12
B. -12/13
C. -5/12
D. 12/13

Answers

Answer 1

Step-by-step explanation:

In Quadrant II, cosine is negative.

We know that sin²x + cos²x = 1, so:

cosx = sqrt[1 - sin²x]

= sqrt[1 - (5/13)²]

= sqrt(144/169)

= -12/13 (since cosx is negative here)

The answer is -12/13. (B)

Answer 2

Option B is correct.

Quadrants and the "cast" Rule:In the first quadrant, the values for sin, cos, and tan are positive.In the second quadrant, the values for sin are positive only.In the third quadrant, the values for tan are positive only.In the fourth quadrant, the values for cos are positive only.

According to the given question

We have,  sin ∅ = 5/13

From the trigonometric identities we know that

[tex]sin^{2}[/tex]∅ + [tex]cos^{2}[/tex]∅ =1

substitute the value of sin∅ in the above identity

[tex](\frac{5}{13} )^{2} + cos^{2}[/tex]∅  =1

⇒ [tex]\frac{25}{69} +cos^{2}[/tex]∅  =1

⇒ [tex]cos^{2}[/tex]∅ = 1 - [tex]\frac{25}{169}[/tex]

⇒ [tex]cos^{2}[/tex]∅ = [tex]\frac{169-25}{69} =\frac{144}{169}[/tex]

⇒[tex]cos[/tex]∅ = [tex]\sqrt{\frac{144}{169} }[/tex]

⇒cos∅ = ±[tex]\frac{12}{13}[/tex]

Since, ∅ is in quadrant II, then cos<0

⇒ cos∅ = -[tex]\frac{12}{13}[/tex]

Hence, option B is correct.

Learn more about quadrant and cast rule here:

https://brainly.com/question/11099638

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Step-by-step explanation:

The centre of the circle is at the midpoint of the diameter.

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