Question

NO LINKS!! Please help me​

Answer 1

Answer:

A

Step-by-step explanation:

$\cos(\theta)=\dfrac{\textsf{adjacent side}}{\textsf{hypotenuse}}=\dfrac{5}{13}$

$\textsf{As }\cos(\theta) > 0 \textsf{ the angle is in quadrant I or IV}$

Using Pythagoras' Theorem a² + b² = c² to find the side opposite the angle:

⇒ 5² + b² = 13²

⇒ b² = 144

⇒ b = 12

⇒ opposite side = 12

$\implies \sin(\theta)=\dfrac{\textsf{opposite side}}{\textsf{hypotenuse}}=\dfrac{12}{13}$

$\textsf{As }\sin(\theta) < 0 \textsf{ then }\sin(\theta)=-\dfrac{12}{13} \textsf{ and the angle is in either quadrant III or quadrant IV}$

Therefore, the common quadrant is quadrant IV and

$\sin(\theta)=-\dfrac{12}{13}$

Answer 2

Answer:

A

Step-by-step explanation:

Since cosine is positive and sine is negative that puts θ in Quad IV.

From right triangles we know:

Cos θ = adjacent/hypotenuse = 5/13

sin θ = opposite/hypotenuse = ?/13

To find the opposite side across from θ use the pythagorean theorem.

5² + y² = 13²

25 + y² = 169

y² = 144

y = 12

we are given  that sin is < 0 so sinθ = -12/13