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1.) What is the value of cosθ given that (−5, −4) is a point on the terminal side of θ?

4√41 / 41

−4√41 / 41

−5√41 / 41

5√41 / 41

2.) What is the value of sinθ given that (−6, −8) is a point on the terminal side of θ?


−3/5

4/5

3/5

−4/5

ONLY ANSWER IF YOU CAN EXPLAIN THIS TO ME CORRECTLY.


Sagot :

Answer:

1) C

2) D

Step-by-step explanation:

1)

We want to find the value of cos(θ) given that (-5, -4) is a point on the terminal side of θ.

This is represented by the diagram below.

Recall that cosine is the ratio of the adjacent side to the hypotenuse.

Find the hypotenuse:

[tex]\displaystlye \begin{aligned} a^2 + b^2 &= c^2 \\ c &= \sqrt{a^2 + b^2} \\ &= \sqrt{(4)^2 + (5)^2} \\ &= \sqrt{41}\end{aligned}[/tex]

The adjacent side with respect to θ is -5 and the hypotenuse is √41. Hence:

[tex]\displaystyle \begin{aligned}\cos \theta &= \frac{(-5)}{(\sqrt{41})} \\ \\ &= - \frac{5\sqrt{41}}{41} \end{aligned}[/tex]

Our answer is C.

2)

We want to find the value of sin(θ) given that (-6, -8) is a point on the terminal side of θ.

This is represented in the diagram below.

Likewise, find the hypotenuse:

[tex]\displaystyle c = \sqrt{(-6)^2 + (-8)^2} = 10[/tex]

Sine is given by the ratio of the opposite side to the hypotenuse.

The opposite side with respect to θ is -8 and the hypotenuse is 10. Hence:

[tex]\displaystyle \begin{aligned}\sin \theta &= \frac{(-8)}{(10)} \\ \\ &= -\frac{4}{5} \end{aligned}[/tex]

Our answer is D.

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