Chapter 2 - Acute Angles and Right Triangles - Section 2.5 Further Applications of Right Triangles - 2.5 Exercises - Page 88: 33

$x\approx 5.32 m$

$\tan 18^{\circ} 10' =\frac{h}{28}$ $\tan 27^{\circ} 10' =\frac{x+h}{28}$ $28\tan 18^{\circ} 10' =h$ $x = 28\tan 27^{\circ} 10' - h$ $x = 28\tan 27^{\circ} 10' - 28\tan 18^{\circ} 10'\approx 5.181574856\approx5.32m$