Answer
The distance between the airplane and the mountain is 5.99 km when the second bearing is taken.
Work Step by Step
Let A be the point where the mountain is at a bearing of $24.1^{\circ}$
Let B be the point where the mountain is at a bearing of $32.7^{\circ}$
Let point C be the location of the mountain.
The points ABC form a triangle.
The angle $A = 24.1^{\circ}$
The angle $B = 180^{\circ}-32.7^{\circ} = 147.3^{\circ}$
We can use the law of sines to find the distance $BC$, which is the distance from the mountain when the second bearing is taken:
$\frac{BC}{sin~24.1^{\circ}} = \frac{7.92}{sin~147.3^{\circ}}$
$BC = \frac{7.92~sin~24.1^{\circ}}{sin~147.3^{\circ}}$
$BC = 5.99~km$
The distance between the airplane and the mountain is 5.99 km when the second bearing is taken.
