Answer
The distance from the aircraft to the enemy headquarters is $\approx 8812$ m.
Work Step by Step
1. Find the angles in the left $\triangle$
$\angle C = 90 - 53$
$= 37^{\circ}$
$\angle A = 180 - (37 + 90)$
$= 180 - 127$
$= 53^{\circ}$
2. Find the angles in the right $\triangle$
$\angle C = 90 - 62$
$= 28^{\circ}$
$\angle B = 180 - (28 + 90)$
$= 180 - 118$
$= 62^{\circ}$
3. Add the $\angle C$ from both triangles to find $\angle C$ of the combined triangle $\triangle ABC$
$= 37 + 28$
$= 65^{\circ}$
4. Use the Law of Sines to find the distance from $C$ to $B$
Let $x = \overline{CB}$
$\frac{x}{sin(53)} = \frac{10,000}{sin(65)}$
$x = \frac{10,000sin(53)}{sin(65)}$
by GDC / calculator
$x = 8811.96...$ m
$x \approx 8812$ m
