Question

Question is posted below

Answer 1

Answer:

-120/169

Step-by-step explanation:

$\sin{\theta} = \pm\sqrt{1-\cos^2{\theta}}=\pm\sqrt{1-(\frac{5}{13})^2}=\pm\sqrt{1-\frac{25}{169}}=\pm\sqrt{\frac{144}{169}}=\pm\frac{12}{13}$

In Quadrant IV, sinθ < 0

so sinθ = -12/13

sin2θ = 2 sinθ cosθ = 2 (5/13) (-12/13) = -120/169