Pls help !!!!!!!!!!!!

Question

asked Dec 25, 2024 186k views

1 Answer

Rich Miller · Answer 1 · 2024-12-30T06:05:45+0000

Answer:

$C.\quad (√(17))/(6)$

Explanation:

We are given

$\tan\theta = - (√(19))/(√(17)) \\\\$

Square both sides

$\tan^2\theta = \left(- (√(19))/(√(17)) \right)^2 = (19)/(17)$

$\tan \theta = (\sin\theta)/(\cos\theta)\\\\\tan^2 \theta = (\sin^2\theta)/(\cos^2\theta)\\$

Substituting for
$tan^2 \theta$ and switching sides we get

$(\sin^2\theta)/(\cos^2\theta) = (19)/(17)\\\\\text{Add 1 to both sides:}\\\\ \\ (\sin^2\theta)/(\cos^2\theta) + 1 = (19)/(17) + 1 \\\$

$(\sin^2\theta + cos^2\theta)/(\cos^2\theta) = (19 + 17)/(17) = (36)/(17)$

$\sin^2\theta + cos^2\theta = 1$

giving

$(1)/(\cos^2\theta) = (36)/(17)$

$(17)/(36) = \cos^2\theta\\\\ \cos^2\theta = (17)/(36)\\\\ \cos\theta = \sqrt{(17)/(36)}\\\\ \cos\theta = (√(17))/(6) \\\\$

This is choice C. You had it right