# Is it possible to solve $\sin(\alpha)$ from right triangle when one side $\sqrt{5-x^2}$, second side $x$, hypotenuse unknown?

So, is it possible to solve $\sin(\alpha)$ from right triangle when one side $\sqrt{(5-x^2)}$, second side $x$, hypotenuse unknown? From my understanding it's not but I don't know these things so well... I just know that $\sin(\alpha)=\frac{\sqrt{(5-x^2)}}{x}$. Thanks for any advice :). (There is a picture about the triangle below as a link.)

• Hint: the hypotenuse is 5, by the Pythagorean theorem. – Parcly Taxel Oct 29 '16 at 12:36
• If you would like more steps for my answer feel free to comment OP. – EnlightenedFunky Oct 29 '16 at 12:42

You could use Pythagoras Theorem:

$$a^2+b^2=c^2$$ $$x^2+5-x^2=c^2$$

Then the $x^2$ would cancel leaving

$$\sqrt5=c$$

So then:

$$\sin(\alpha)=\frac{\sqrt{5-x^2}}{\sqrt5}=\frac{\sqrt{25-5x^2}}{5}$$

USING PYTHAGORAS THEOREM WE CAN GET THE HYPOTENUSE

$hypotenuse^2=x^2+5-x^2$

so $hypotenuse$ becomes $\sqrt5$

*WE KNOW *

$\sin(\alpha)=perpendicular/hypotenuse$

$\sin(\alpha)=\frac{\sqrt{5-x^2}}{\sqrt5}=\frac{\sqrt{25-5x^2}}{5}$

$\sin(\alpha)=\frac{\sqrt{(5-x^2)}}{x}$

PROOVED

It is not possible if you don't know the value of $x$ (I recall the notation $x$ is usually for unknowns). In fact the hypotenuse is clearly equal to $\sqrt 5$ but there are infinitely many right triangles having a given hypotenuse $a$ (in the circle of diameter $a$ all these triangles have the vertex of $90^{\circ}$ in the corresponding circumference). Easily you have directly from your figure $$\sin(\alpha)=\sqrt{\frac{5-x^2}{5}}$$