Hello from MrBillDoesMath
Answer:
cos@ = - sqrt(7)/ 4
which is choice B
Discussion:
This problem can be solved by drawing triangles and looking at ratios of sides or by using the trig identity:
(cos@)^2 + (sin2)^2 = 1
If sin@ = 3/4 , the
(cos@)^2 + (3/4)^2 = 1 => (subtract (3/4)^2 from both sides)
(cos@)^2 = 1 - (3/4)^2 = 1 - 9/16 = 7/16
So...... taking the square root of both sides gives
cos@ = +\- sqrt(7)/ sqrt(16) = +\- sqrt(7)/4
But is cos@ positive or negative? We are told that @ is in the second quadrant and cos(@) is negative in this quadrant, so our answer must be negative
cos@ = - sqrt(7)/ 4
which is choice B
Thank you,
Mr. B
The value of [tex]cos\theta=\frac{\sqrt7}{4}[/tex]
In Quadrant II, only [tex]sin\theta[/tex] is positive. That is,
[tex]sin\theta\text{ is positive}\\cos\theta\text{ is negative}[/tex]
To find the value of [tex]cos\theta[/tex] from the given value of [tex]sin\theta[/tex], note that the triangle the ratios are based on have the sides [tex]x,y,r[/tex], where
[tex]x=\text{the side adjacent the reference angle of }\theta\text{ in Quadrant II}\\y=\text{the side opposite the reference angle of }\theta\text{ in Quadrant II}\\r=\text{the hypotenuse}[/tex]
so, the [tex]sin\theta[/tex] and [tex]cos\theta[/tex] ratios will be given by
[tex]sin\theta=\frac{y}{r}=\frac{3}{4}[/tex]
[tex]cos\theta=\frac{x}{r}=\frac{x}{4}[/tex]
to find [tex]x[/tex], we will need to use Pythagoras' theorem
[tex]x=\sqrt{r^2-y^2}\\x=\sqrt{4^2-3^2}\\=\sqrt{16-9}=\sqrt7[/tex]
since [tex]cos\theta=\frac{x}{r}[/tex], we will have
[tex]cos\theta=\frac{\sqrt7}{4}[/tex]
Learn more about trigonometry here: https://brainly.com/question/95152
