If an angle is \(10^\circ\) more than twice its complement, what is the angle?

Let the angle be $x^\circ$.
Its complement is $(90 - x)^\circ$.

According to the question, the angle is $10^\circ$ more than twice its complement.

So,
$x = 2(90 - x) + 10$

$x = 180 - 2x + 10$

$x = 190 - 2x$

$3x = 190$

$x = \frac{190}{3} = 63\frac{1}{3}^\circ$

Thus, the angle is approximately $63^\circ$.

Answer = $63^\circ, 33^\circ$.