The angle is defined at quadrant 1 and it's given that tanx=(11)^(1/2)/5
Find cosx=
cscx=
Hi HMV,
$\displaystyle \tan X=\frac{\sqrt{11}}{5}$
We know that $\displaystyle tan X = \frac{y}{x}$
That means $\displaystyle x = 5$ and $\displaystyle y = \sqrt{11}$
Now use $\displaystyle r^2=x^2 + y^2$ to solve for r.
Once you have values for x, y, and r you can substitute them into the following:
$\displaystyle \cos X = \frac{x}{r}$
$\displaystyle \csc X = \frac{r}{y}$
Hello HMV$\displaystyle \tan x = \tfrac{\sqrt{11}}{5}$
Since the angle is in the first quadrant, it's an acute angle. So you can draw a right-angled triangle where the opposite side to angle $\displaystyle x$ is $\displaystyle \sqrt{11}$ and the adjacent $\displaystyle = 5$.
So the $\displaystyle \text{hypotenuse}^2 = 11 + 25 = 36$
$\displaystyle \Rightarrow \text{hypotenuse} = 6$
Can you complete it now?
Grandad