Which are the roots of the quadratic function f(q) = q2 – 125? Select two options.
2 answers:
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Answer:
measure of angle UMN is 160.
hope u got it...
Answer:
So 67 degrees is one value that we can take the sine of such that is equal to cos(23 degrees).
(There are more values since we can go around the circle from 67 degrees numerous times.)
Step-by-step explanation:
You can use a co-function identity.
The co-function of sine is cosine just like the co-function of cosine is sine.
Notice that cosine is co-(sine).
Anyways co-functions have this identity:
or
You can prove those drawing a right triangle.
I drew a triangle in my picture just so I can have something to reference proving both of the identities I just wrote:
The sum of the angles is 180.
So 90+x+(missing angle)=180.
Let's solve for the missing angle.
Subtract 90 on both sides:
x+(missing angle)=90
Subtract x on both sides:
(missing angle)=90-x.
So the missing angle has measurement (90-x).
So cos(90-x)=a/c
and sin(x)=a/c.
Since cos(90-x) and sin(x) have the same value of a/c, then one can conclude that cos(90-x)=sin(x).
We can do this also for cos(x) and sin(90-x).
cos(x)=b/c
sin(90-x)=b/c
This means sin(90-x)=cos(x).
So back to the problem:
cos(23)=sin(90-23)
cos(23)=sin(67)
So 67 degrees is one value that we can take the sine of such that is equal to cos(23 degrees).
