The following equation reflects the value of y:
[tex]\to y=\sqrt{x^2-w^2}[/tex]
In terms of triangle BCD, we have us the Pythagoras theorem:
[tex]\to BC^2=CD^2+BD^2[/tex]
Values for BC, CD, and BD can be substituted.
[tex]\to x^2=w^2+y^2[/tex]
Subtraction of [tex]w^2[/tex] from both sides
[tex]\to x^2-w^2=y^2[/tex]
Calculate the square roots of both sides.
[tex]\to \sqrt{x^2-w^2}=y[/tex]
Rephrase as:
[tex]\to y=\sqrt{x^2-w^2}[/tex]
As a result, the equation representing the value of y is: [tex]y=\sqrt{x^2-w^2}[/tex]
Learn more about the right-angled triangle:
brainly.com/question/3770177