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Nikitich [7]
3 years ago
11

What is the exterior angle of triangle ABC These are the points: A=3, B=10, C=6

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
3 0
Based off of  what I can see in the  photo (its too blurry to know for sure) and what I think you're asking:

The angle of a triangle always adds up to 360degrees, so base it on that and the acute and obtuse  angles  and get the degrees from there, the angle would be those degrees you find or <10, 5, 4 (what I assume  is a 5)

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The cosine of 23° is equivalent to the sine of what angle
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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:

\cos(90^\circ-x)=\sin(x)

or

\sin(90^\circ-x)=\cos(x)

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:

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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).

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