Question

If cos theta = -3/5 in quadrant 2, what is sin theta?

Answer 1

Answer:

hello : sin(θ) = 4/25

Step-by-step explanation:

you know : cos²(θ) + sin²(θ) =1   and : cos(θ) = - 3/5

(-3/5)² + sin²(θ) =1

9/25 + sin²(θ) =1

sin²(θ) =1 -9/25

sin²(θ) = 16/25

sin(θ) = 4/25 or sin(θ) = - 4/25

in quadrant 2 :  sin(θ) > 0

so : sin(θ) = 4/25

Answer 2

Answer:

$sin(theta)=\frac{4}{5}$

Step-by-step explanation:

if cos theta = -3/5 in quadrant 2, find out sin(theta)

$cos(x)=\frac{-3}{5}$

cos is negative in second and third quadrant

Here cos is in second quadrant.

Sin is positive in second quadrant .

$cos(x)=\frac{adjacent}{hypotenuse}$

adjacent side= 3 , hypotenuse = 5

Lets use Pythagorean theorem $c^2=a^2 + b^2$

$5^2=3^2 + b^2$

$25=9 + b^2$ Subtract 9 from both sides

$b^2= 16$

b=4

opposite side = 4

$sin(x)=\frac{opposite}{hypotenuse}$

$sin(theta)=\frac{4}{5}$