Use the triangle to find each of the six trigonometric functions ...

Question

Use the triangle to find each of the six trigonometric functions of θ.

Right triangle with sides labeled 4 and 11, angle α, and a right angle at point O.

Accepted Answer

Hello, everyone. We are asked to refer to the triangle shown and determine the value of each of the six trigonometric functions of alpha sine alpha, cosine alpha, tangent alpha, cotangent alpha, C K, alpha and co C can alpha, we have a right triangle O N M where the right angle is at angle N alpha is at M side, O N measures four side M N measures 11. So right now, I know I don't have my hypotenuse as measure. So I'm gonna call that X and before I do anything else, I'm gonna use the Pythagorean to find the value of X. So recall the Pythagorean theorem is the sum of the squares of the legs of a right triangle are equal to the square of its hypotenuses. So in this case, four squared plus 11 squared is gonna equal my X or my hypotony squared. Fourth grade is 16, 11 squared is 121 that's gonna equal X squared. Adding these together. I get 137 equals X squared. I'm gonna take the square root of both sides and I'm gonna treat X as the square root of 137 So that is my hypotenuse the square root of 137. So now going through all six trig functions, I will rationalize the denominators. If need be, I'm gonna start with sign, the sign of alpha would be the opposite side divided by the hypotenuse. So the site opposite alpha is O N which has a value of four and then divided by the hypotenuse which is radical 137. Again, we can't leave that like that. So we're going to rationalize it by multiplying both the numerator and the denominator by radical 137. And we get that the sign of alpha is four radical divided by 137. Next, the cosine of alpha is the adjacent side divided by the hypotenuse. So the site adjacent to alpha has a measure of 11. Our hypotenuse is the square root of 137. And just like with the cosine, we don't want to leave a square root in the denominator. So we're rationalizing it by multiplying numerator and the denominator by the square root of 137 which gives us the value multiplied by radical 137 divided by 137. Next, we have tangent. So tangent of alpha, it's gonna be the opposite side divided by the adjacent side. So the opposite side is four, the adjacent side is 11. Nothing to rationalize there. 4/11. Next, we have the of alpha. And I know that that is the opposite of the sign. So that is the hypotenuse divided by the opposite side. So the hypotenuse is the square root of divided by the side opposite it, which is O N which has a value of four. Since the radicals in the numerator, no need to rationalize it. Next, the sea can of alpha which again is the reciprocal of cosine. So this would be the hypotenuse divided by the adjacent side. So radical 137 divided by 11 and again, no need to rationalize that as the radicals in the numerator and then the cotangent of alpha is the reciprocal of the tangent. So it's gonna be the adjacent side divided by the opposite side. So this would be 11 divided by four. So we have all six trig ratios. Sine is four radical 137 divided by cosine is 11 radical 137 divided by 137. Tangent is 4/11. Koi Kent is radical 137 divided by four C K is radical 137 divided by 11 and cotangent is 11 4th have a nice day.

Use the triangle to find each of the six trigonometric functions of θ.

Key Concepts

Trigonometric Functions

Pythagorean Theorem

Angle of Elevation

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