Mathematics: NO LINKS!! Please help me

C. equidistant from the endpoints of the segment...

NO LINKS!! Please help me - id15357471, jacamron

DestroYa121

15.03.2022

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

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

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