Solution :
Let the numbers are a and $b$.
$\therefore$ According to question,
and $a^{2}+b^{2}=289$
We know that, $$ \begin{aligned}(a+b)^{2} &=a^{2}+b^{2}+2 a b \\ &=289+2 \times 120 \\ &=289+240=529 \end{aligned} $$ $\therefore \quad a+b=\sqrt{529}=23 ;$ Ans.
$$
\begin{aligned}
(a+b)^{2} &=a^{2}+b^{2}+2 a b \\
&=289+2 \times 120 \\
&=289+240=529 \\
\therefore a+b=& \sqrt{529}=23 ; \text { Ans. }
\end{aligned}
$$
Trick :
$$
\begin{aligned}
a+b &=\sqrt{289+2 \times 120} \\
&=\sqrt{529}=23: \text { Ans. }
\end{aligned}
$$
