Given:
ZY=90 ft, YX=672 ft, and ZX=678 ft.
WY=t
The line segment WY is perpendicular to the side ZX from the third vertex Z.
So WY is the altitude of the given triangle.
Let a=90 ft, b=672 ft and c=678 ft, and t is altutude.
The formula for the altitude is
[tex]t=\frac{2\sqrt[]{s(s-a)(s-b)(s-c)}}{b}[/tex]
Here s is the semiperimeter of the tirangle .
[tex]s=\frac{a+b+c}{2}[/tex]
Substitute a=90ft, b=672, and c=678 in the equation , we get
[tex]s=\frac{90+672+678}{2}=720\text{ ft}[/tex]
We get s=720ft.
Substitute s=720ft, a=90ft, b=672, and c=678 in the altitute formula, we get
[tex]t=\frac{2\sqrt[]{720(720-90)(720-672)(720-678)}}{672}[/tex]
[tex]t=\frac{2\sqrt[]{720\times630\times\times48\times42}}{672}[/tex]
[tex]t=\frac{2\sqrt[]{914457600}}{672}=90[/tex]
Hence the value of the variable is 90 ft.
