^{2}+b

^{2}=c

^{2}. a and b = 4 and 11, in either order, according to the problem. Therefore, c

^{2}=137; c=sqrt137.

Thanks for the help! If you could please show all your work? Thanks again!

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Kirill and Felice gave good answers, but I wanted to add this to eliminate any confusion.

We can only use trig functions with right triangles, so you have to use Pythagorean's Theorem to solve for the hypotenuse: a^{2}+b^{2}=c^{2}. a and b = 4 and 11, in either order, according to the problem. Therefore, c^{2}=137; c=sqrt137.

I hope that fills in any gaps. Kirill and Felice covered the rest of the explanation.

Since sec(x)<0 and tan(x)>0, angle is in the third quadrant.

cot(x)=11/4

cos(x)=-√[1/(1+tan^{2}(x))]=-√[1/(1+16/121)]=-11/√137;

sin(x)=-√[1-cos^{2}(x)]=-√[1-121/137]=-4/√137

sec(x)=-(√137)/11

cosec(x)=-(√137)/4

Well if we use the trigonometric functions

sin X = opposite / hypotenuse

cos X = adjacent / hypotenuse

tan X = opposite/ adjacent

csc X = hypotenuse / opposite

sec X = hypotenuse/adjacent

ctn X = adjacent/opposite

We know that the hypotenuse = sqrt(adjacent^{2 }+ opposite^{2})^{
}

from your given tan X = 4/11 we could say the the opposite was 4 and adjacent was 11

so our hypotenuse would be sqrt(11^{2} + 4^{2}) = sqrt (121+16) = sqrt(137) = 11.7

so your sin X = opp/hyp = 4/11.7

fill in the rest for the other 4 functions

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