Question

The point (−3, 1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.

Answer 1

ANSWER

See below

EXPLANATION

The given point (-3,1) lies in the second quadrant.

In this quadrant only sine is positive.

The length of the hypotenuse formed by the right angle triangle is

${h}^{2} = {3}^{2} + {1}^{2}$

${h}^{2} = 9 + 1$

${h}^{2} = 10$

$h = \sqrt{10}$

The side opposite to θ, is 1 units.

The adjacent side is 3 units.

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

$\sin( \theta) = \frac{1}{ \sqrt{10} } = \frac{ \sqrt{10} }{10}$

$\cos( \theta) = - \frac{adjacent}{hypotenuse}$

$\cos( \theta) = - \frac{3}{ \sqrt{10} } = - \frac{3 \sqrt{10} }{10}$

$\tan( \theta) = - \frac{opposite}{adjacent}$

$\tan( \theta) = - \frac{1}{3}$

Answer 2

Answer:

The angle is in the second quadrant  so only the sine is positive . 

Length of the hypotenuse = sqrt (-1^2 + 3^2)  = sqrt10

sine  =  1/sqrt10 =  0.3162 to 4 dec places

tangent = 1 / -3 =  -0.3333  to 4 d p's

cosine = -3/sqrt10 = -0.9487 to 4 d, p's.