What are the six trigonometric ratios the hypotenuse is x opposite side is 3 adjacent is 3 square root of 10?

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Answer 1 · 2018-04-04T14:04:55

See below.

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First we need to find the value of #x#

Using Pythagors's theorem:

#x^2=3^2+(3sqrt(10))^2#

#x^2=9+90=99#

#x=sqrt(99)=3sqrt(11)#

The six ratios are:

#sin(theta)="opposite"/"hypotenuse"=3/(3sqrt(11))=color(blue)(sqrt(11)/11)#

#cos(theta)="adjacent"/"hypotenuse"=(3sqrt(10))/(3sqrt(11))=color(blue)(sqrt(110)/11)#

#tan(theta)="opposite"/"adjacent"=3/(3sqrt(10))=color(blue)(sqrt(10)/10)#

#csc(theta)="hypotenuse"/"opposite"=1/sin(theta)=1/(sqrt(11)/11)=color(blue)(sqrt(11)#

#sec(theta)="hypotenuse"/"adjacent"=1/cos(theta)=1/(sqrt(110)/11)=color(blue)(sqrt(110)/10)#

#cot(theta)="adjacent"/"opposite"=1/tan(theta)=1/(sqrt(10)/10)=color(blue)(sqrt(10)#