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OpenStudy (anonymous):

Problem: find the measure of the angle to the nearest minute if cosB=0.8857 I tried: cos(0.8857) = .6327 but cos(.6327=.8064 in radian mode. cos(.8857)=.9999 and cos(.9999)= .9998 in degree mode. I am lost, can you help?

OpenStudy (amistre64):

so find the measure of an angle when you know its ratio; use the inverse trig functions. cos^-1(.8857)

OpenStudy (anonymous):

Thank you, thank you. Could you explain why that works?

OpenStudy (amistre64):

the technical is is simply because the trig functions are defined by rations of the sides of a rt triangle. cosine(angle) = \(side.net.to\over hypotenuse\) by definition. Since all the ratio for the trig functions are unique in the domain from 0 to 2pi, if we know a ration, we can determine the angle which it is associated with

OpenStudy (amistre64):

\[cos(\theta)=\frac{side.adjacent.to.it}{hyporenuse}\]

OpenStudy (amistre64):

lol .... i cant spell tonight

OpenStudy (amistre64):

\[cos(Angle)=x/r\] \[Angle = cos^{-1}(x/r)\] is all it means

OpenStudy (amistre64):

im sure there are some technical fauxpauxs in that; since the domain for the cosine inverse is from 0 to pi :)

OpenStudy (anonymous):

Does this mean that cos(angle) can be used as factors such as 2(3) = x/r can be manipulated to 3 = 1/2(X/r)??

OpenStudy (amistre64):

if i read this right.... not quite; 'cos' is simply the name of a function; it is akin to an operator on an angle to come up with the value of the angle in terms of the adjacent length/hypotenuse. but then, I dont quite understand your reply

OpenStudy (amistre64):

the term 'cos(45)' itself is not a term of a product of 'cos' times 45

OpenStudy (amistre64):

cos(45) = \(1/\sqrt{2}\) cannot be algebraically manipulated as: 45 = \(1/cos(\sqrt{2})\)

OpenStudy (amistre64):

cos(45) means that the ration of a triangle fromed by a 45 degree angle creates a baseleg of 1, and a height leg of 1; and a hypotenuse of sqrt(2); the cos(45) equals the baseleg of 1 divided by the hypotenuse of sqrt(2).

OpenStudy (amistre64):

lol .... i dint use the unit circle there, but it still amounts to the same value

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