Prompt
Explain, in your own words, how to find the values of the six trigonometric ratios.

There are six trigonometric ratios defined in the right triangles. Below you will find their definitions and how to calculate them.

Explanation:

The six trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent.

The symbols used for them are:

sine: sin

cosine: cos

tangent: tan

cosecant: csc

secant: sec

cotangent: cot

The trigonometric ratios are defined as the ratio of the sides in right triangles.

In a right triangle the two sides opposite to the accute angles are called legs, and the side opposite to the right angle (the larger side) is called hypotenuse.

The sine of an angle is the ratio of the opposite leg to the angle to the hypotenuse. So, you find the sine by doing the quotient of the length of the opposite leg to the angle and the length of the hypotenuse.

sine (angle) = length of the opposite leg / length of the hypotenuse.

The cosine of an angle is the ratio of the adjacent leg to the angle to the hypotenuse. So, you find the cosine by doing the respective division:

cosine (angle) = length of the adjacent leg / length of the hypotenuse

The tangent of an angle is the ratio of the sine to the cosine of the same angle. So, you find it either by dividing sine by cosine or by dividing the length of the opposite angle by the length of the adjacent angle.

tangent (angle) = sine (angle)/ cosine (angle) = opposite leg / adjacent leg.

Cosecant is the inverse of the sine, secant is the invers of the cosine, and cotangent is the inverse of the tangent. So, you find them by these equations:

cosecant (angle) = 1 / sine (angle) = hypotenuse / opposite leg

secant (angle) = 1 / cosine (angle) = hypotenuse / adjacent leg

cotangent (angle) = 1 tangent (angle) = adjacent leg / opposite leg.

horizontal translation of 2 units in the positive x direction

followed by

vertical translation of 3 units in the positive y direction

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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