Use the Pythagorean theorem since you are working with a right triangle:
a^2+b^2=c^2a2+b2=c2
The legs are a and b and the hypotenuse is c. The hypotenuse is always opposite the 90° angle. Insert the appropriate values:
0.8^2+0.6^2=c^20.82+0.62=c2
Solve for c. Simplify the exponents (x^2=x*xx2=x∗x ):
0.64+0.36=c^20.64+0.36=c2
Add:
1=c^21=c2
Isolate c. Find the square root of both sides:
\begin{gathered}\sqrt{1}=\sqrt{c^2}\\\\\sqrt{1}=c\end{gathered}1=c21=c
Simplify \sqrt{1}1 . Any root of 1 is 1:
c=c= ±11 *
c=1,-1c=1,−1
