#### Answer

$5\sqrt{3} \text{ and } 10\sqrt{3}$

#### Work Step by Step

Since the given acute angle of the right triangle is $60^o,$ then the side opposite this angle, $a,$ measures $\dfrac{\sqrt{3}}{2}\cdot c,$ where $c$ is the hypotenuse. Hence, the hypotenuse is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{3}}{2}\cdot c=a
\\\\
c=15\cdot\dfrac{2}{\sqrt{3}}
\\\\
c=15\cdot\dfrac{2}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}
\\\\
c=15\cdot\dfrac{2\sqrt{3}}{3}
\\\\
c=10\sqrt{3}
.\end{array}
Let the right triangle have $a$ and $b$ as the legs and $c$ as the hypotenuse. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, with $
a=15
$ and $
c=10\sqrt{3}
,$ then
\begin{array}{l}\require{cancel}
a^2+b^2=c^2
\\\\
15^2+b^2=(10\sqrt{3})^2
\\\\
\\\\
225+b^2=100(3)
\\\\
225+b^2=300
\\\\
b^2=300-225
\\\\
b^2=75
\\\\
b=\sqrt{75}
\\\\
b=\sqrt{25\cdot3}
\\\\
b=\sqrt{(5)^2\cdot3}
\\\\
b=5\sqrt{3}
.\end{array}
Hence, the missing sides have measure $
5\sqrt{3} \text{ and } 10\sqrt{3}
$ units.