Question

I need help on this!
Given the following right triangle △ ABC with m∠A = Θ°, find sin(Θ°) and cos(Θ°).

Answer 1

Answer:

$\sf Sin \ \theta = \dfrac{5}{13}\\\\Cos \ \theta =\dfrac{12}{13}$

Step-by-step explanation:

  $\sf Sin \ \theta= \dfrac{opposite \ of \ angle \ A}{hypotenuse}$

            $\sf =\dfrac{BC}{AB}\\\\ =\dfrac{5}{13}$

$\sf Cos \ \theta = \dfrac{adjacent \ side \ of \ angle \ A}{hypotenuse}$

       $\sf =\dfrac{AC}{AB}\\\\= \dfrac{12}{13}$