Question

Consider this equation. if is an angle in quadrant iv, what is the value of ? a. b. c. d.

Answer 1

Option (a) is correct. The value of sinθ will be $-\frac{5\sqrt{41} }{41}$

We are given the value of the cosθ  = $\frac{4\sqrt{41} }{41}$ where, θ is in the IV Quadrant.

We know that cosine is the ratio of the adjacent side to the hypotenuse.  Also, we have
hypotenuse = h = 41, And Adjacent Side = b = $4\sqrt{41}$
So, for the value of the opposite side, we need to use the Pythagorean theorem which is stated as:$(Hypotenuse)^{2} = (opposite side)^{2} + (adjacentside)^{2}$
$h^{2} = a^{2} + b^{2}$

$a^{2} = h^{2} - b^{2}$

a = $\sqrt{41^{2} -(4\sqrt{41})^{2} }$ =  $\sqrt{1681 - 656 }$ = $\sqrt{1025}$ = $5\sqrt{41}$

The opposite side is  $5\sqrt{41}$ , as sine is negative (-) in the IV quadrant so the value will be   $-5\sqrt{41}$.

So we know that Sine is the ratio of the Opposite side to the hypotenuse, which can be written as:

sineθ = $\frac{Opposite}{Hypotenuse}$

sineθ =  $-\frac{5\sqrt{41} }{41}$

So, option (a) will be the right option.

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Your Question is incomplete, but most probably your full question was

Consider this equation. cosθ  = $\frac{4\sqrt{41} }{41}$ if θ is an angle in quadrant IV, what is the value of sinθ?
(a)       $-\frac{5\sqrt{41} }{41}$

(b)   $\frac{5\sqrt{41} }{41}$

(c) $-\frac{5}{4}$

(d) $\frac{5}{4}$