Answer
The surveyor is around $\approx 42.7 $ ft from the base of the building.
Work Step by Step
Looking at Figure #1 in the attachment:
$A =$ Position of surveyor
$B = $ Base of the building
$C = $ Point horizontally across from the surveyor to the building
$D = $ Top of the building
In total, there are two right angled triangles present, $\triangle ABC$ and $\triangle ADC$.
1. Find the angles in all of the triangles
$\triangle ABC$:
$\angle C = 90˚$ (Properties of a right angled triangle)
$\angle BAC = 16˚$ (Given by the question)
$\angle B = [180 - (90 + 16)]$ (Angles in a triangle must add up to $180˚$)
$\angle B = [180 - 106]$
$\angle B = 74˚$
$\triangle ADC$:
$\angle C = 90˚$ (Properties of a right angled triangle)
$\angle DAC = 43˚$ (Given by the question)
$\angle D = [180 - (43 + 90)]$ (Angles in a triangle must add up to $180˚$)
$\angle D = [180 - 133]$
$\angle D = 47˚$
2. Verify that $\angle A$, $\angle B$ and $\angle D$ adds to $180˚$
$\angle A + \angle B+\angle D = 180˚$
$(43 + 16) + (74) + (47) = 180˚$
$59 + 74 + 47 = 180˚$
$180˚ = 180˚$
3. Solve for $x$ using the sine law (Use a calculator)
$\frac{50}{sin59} = \frac{x}{sin47}$
$x = \frac{50sin47}{sin59}$
$x = \frac{36.5676...}{0.8571...}$
$x \approx 42.7$ ft
