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Determine all six trigonometric ratios (sine, cosine, tangent, cosecant, secant, and cotangent) for the triangle pictured.

Determine all six trigonometric ratios (sine, cosine, tangent, cosecant, secant, and-example-1

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Explanation:

imagine the red dot (the triangle vertex with the marked angle theta) is the center of the circle, and the circle passes through the blue dot.

rotate the triangle up a little bit around the red dot, so that 48 is horizontal, and 55 vertical.

then this triangle is a trigonometric triangle.

so, 73 is the radius (going from the center of the circle to the circle arc).

we know sine is the up/down distance of the point on the circle arc from the circle center.

and cosine is the left/right distance of the point on the circle arc from from the circle center.

and then, don't forget : the trigonometric functions represent these distances in a norm circle (radius = 1).

all the corresponding distances in a larger circle and triangle are the basic trigonometric functions multiplied by the actual radius (scaling factor).

so,

55 = sin(theta) × 73

sin(theta) = 55/73 = 0.753424658...

48 = cos(theta) × 73

cos(theta) = 48/73 = 0.657534247...

tangent = sine/cosine

tan(theta) = 55/73 / 48/73 = (55×73)/(73×48) =

= 55/48 = 1.145833333...

cosecant = 1/sine

csc(theta) = 1 / 55/73 = 73/55 = 1.327272727...

secant = 1/cosine

sec(theta) = 1 / 48/73 = 73/48 = 1.520833333...

cotangent = 1/tangent

cot(theta) = 1 / 55/48 = 48/55 = 0.872727273...

as you can see : the trigonometric functions are really that easy.

FYI : using the inverse of these functions we see that

theta = 48.88790956...°

User Gophermofur
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