1. The distance between two pillars is 150 meters. One is three times as tall as the other. From the midpoint of the line connecting the bases of the two pillars, the angles of elevation to their tops are complementary. What is the height of the shorter pillar?

2. The distance between two pillars is 150 meters. One is three times taller than the other. From the midpoint of the line segment connecting the bases of the two pillars, the angles of elevation to the tops of the pillars are complementary. Find the height of the shorter pillar.

3. Two poles are placed 120 meters apart, and the height of one pole is double that of the other. If the angles of elevation to the tops of the two poles from the midpoint of the line joining their bases are complementary, find the height of the shorter pole.

4. Two pillars are 120 meters apart, with one pillar's height being twice that of the other. At the midpoint of the line segment connecting their bases, the angles of elevation to the tops of the pillars are complementary. Determine the heights of the pillars.

5. Here is the English translation: > Two pillars of equal height are located directly opposite each other at points A and B on either side of a 120-meter wide road. From point C, on the line joining their bases, the angles of elevation to the tops of the pillars at A and B are 60° and 30°, respectively. Find the length of AC.

6. Let the smaller pillar be AB \(= x\) meters and the larger pillar be CD \(= 2x\) meters.
The midpoint of the base connection BD is O; the angles of elevation from O to the tops of the pillars are \(\angle\)AOB = \(\theta\) and \(\angle\)COD = 90\(^o-\theta\).
Since O is the midpoint of BD, we have BO = OD = \(\frac{120}{2}\) meters = 60 meters.

From \( \triangle \)ABO, we get:
\(\cfrac{AB}{BO} = \tan \theta\)
Or, \(\cfrac{x}{60} = \tan\theta\) --------(i)

From \( \triangle \)COD, we get:
\(\cfrac{CD}{OD} = \tan(90^o - \theta)\)
Or, \(\cfrac{2x}{60} = \cot\theta\) --------(ii)

Multiplying equations (i) and (ii), we get:
\(\cfrac{x}{60} \times \cfrac{2x}{60} = \tan\theta \times \cot\theta\)
Or, \(\cfrac{2x^2}{60 \times 60} = 1\)
Or, \(x^2 = \cfrac{60 \times \cancel{60}30}{\cancel{2}}\)
Or, \(x = 30\sqrt2\)

\(\therefore\) The height of the smaller pillar is \(30\sqrt2\) meters.
And the height of the larger pillar is \(30\sqrt2 \times 2\) meters \(= 60\sqrt2\) meters.
(Proved).

7. Two points on the ground lie along the same straight line with the base of a vertical pillar. From these two points, the angles of elevation to the top of the pillar are complementary. If the distances from the two points to the base of the pillar are 9 meters and 16 meters respectively, and both points are on the same side of the pillar, find the height of the pillar.

8. If the angles of depression from the top of a lighthouse to the bases of the masts of two ships situated along the same straight line are 60° and 30° respectively, and the nearer ship is 150 meters away from the lighthouse, then calculate the distance of the farther ship from the lighthouse and the height of the lighthouse.

9. From two points located on the same side and along the same horizontal line passing through the base of a vertical pillar, the angles of elevation to the top of the pillar are respectively \(\theta\) and \(\phi\). If the height of the pillar is \(h\), find the distance between the two points.