Method 1
Applying the Pythagorean Theorem
we know that
[tex]10^{2}= 5^{2} +a^{2}[/tex]
Solve for a
[tex]100= 25 +a^{2}[/tex]
[tex]a^{2}=100-25[/tex]
[tex]a^{2}=75[/tex]
[tex]a=\sqrt{75}=5 \sqrt{3}\ units[/tex]
therefore
the answer is
the length of the altitude is [tex]5 \sqrt{3}\ units[/tex]
Method 2
we know that
[tex]sin(60\°)=\frac{\sqrt{3}}{2}[/tex] -------> equation A
and
in this problem
[tex]sin(60\°)=\frac{a}{10}[/tex] --------> equation B
equate equation A and equation B
[tex]\frac{\sqrt{3}}{2}=\frac{a}{10}\\\\a=\frac{10\sqrt{3}}{2}\\\\a=5 \sqrt{3}\ units[/tex]
therefore
the answer is
the length of the altitude is [tex]5 \sqrt{3}\ units[/tex]
