Solution:

Let the height of the tower be \( h \) meters and the width of the road be \( x \) meters.

From point A (directly opposite):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]

From point B (10m from A towards the tower):

\[ \tan 30^\circ = \frac{h}{x + 10} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x + 10} \]

Substitute \( h = x\sqrt{3} \):

\[ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 10} \]

\[ x + 10 = 3x \]

\[ 2x = 10 \Rightarrow x = 5 \text{ meters} \]

Then \( h = 5\sqrt{3} \) meters.

Answer: The height of the tower is \( 5\sqrt{3} \) meters and the width of the road is 5 meters.

Solution:

Effective height of temple above boy’s eye level = 30 – 1.5 = 28.5 meters

Let initial distance be \( x \) meters and distance walked be \( d \) meters.

From initial position:

\[ \tan 30^\circ = \frac{28.5}{x} \Rightarrow x = 28.5\sqrt{3} \]

From final position:

\[ \tan 60^\circ = \frac{28.5}{x – d} \Rightarrow x – d = \frac{28.5}{\sqrt{3}} \]

Substitute \( x = 28.5\sqrt{3} \):

\[ 28.5\sqrt{3} – d = \frac{28.5}{\sqrt{3}} \]

\[ d = 28.5\sqrt{3} – \frac{28.5}{\sqrt{3}} \]

\[ d = 28.5\left(\sqrt{3} – \frac{1}{\sqrt{3}}\right) = 28.5\left(\frac{3 – 1}{\sqrt{3}}\right) = \frac{57}{\sqrt{3}} = 19\sqrt{3} \text{ meters} \]

Answer: The boy walked \( 19\sqrt{3} \) meters towards the temple.

Solution:

Let the height of the statue be \( h \) meters and distance from point to pedestal be \( x \) meters.

For pedestal (2m height):

\[ \tan 45^\circ = \frac{2}{x} \Rightarrow x = 2 \text{ meters} \]

For statue (2 + h meters height):

\[ \tan 60^\circ = \frac{2 + h}{2} \Rightarrow \sqrt{3} = \frac{2 + h}{2} \]

\[ 2 + h = 2\sqrt{3} \Rightarrow h = 2\sqrt{3} – 2 = 2(\sqrt{3} – 1) \text{ meters} \]

Answer: The height of the statue is \( 2(\sqrt{3} – 1) \) meters.

Solution:

Let height of building be \( h \) meters and height of tower be \( H \) meters.

From angle of depression (45°):

\[ \tan 45^\circ = \frac{h}{7} \Rightarrow h = 7 \text{ meters} \]

From angle of elevation (60°):

\[ \tan 60^\circ = \frac{H – h}{7} \Rightarrow \sqrt{3} = \frac{H – 7}{7} \]

\[ H – 7 = 7\sqrt{3} \Rightarrow H = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \text{ meters} \]

Answer: The height of the tower is \( 7(1 + \sqrt{3}) \) meters.

Solution:

Let height of pole be \( h \) meters.

Original wire (18m at 30°):

\[ \sin 30^\circ = \frac{h}{18} \Rightarrow h = 18 \times \frac{1}{2} = 9 \text{ meters} \]

New wire (at 60°):

\[ \sin 60^\circ = \frac{9}{L} \Rightarrow L = \frac{9}{\sin 60^\circ} = \frac{9}{\sqrt{3}/2} = \frac{18}{\sqrt{3}} = 6\sqrt{3} \text{ meters} \]

Length cut = Original – New = \( 18 – 6\sqrt{3} = 6(3 – \sqrt{3}) \) meters

Answer: \( 6(3 – \sqrt{3}) \) meters of wire was cut.

Solution:

Let height of building be \( h \) meters and distance between them be \( d \) meters.

From building’s foot to tower’s top (30m):

\[ \tan 60^\circ = \frac{30}{d} \Rightarrow d = \frac{30}{\sqrt{3}} = 10\sqrt{3} \text{ meters} \]

From tower’s foot to building’s top:

\[ \tan 30^\circ = \frac{h}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{10\sqrt{3}} \]

\[ h = \frac{10\sqrt{3}}{\sqrt{3}} = 10 \text{ meters} \]

Answer: The height of the building is 10 meters.

Solution:

Let height of poles be \( h \) feet.

Let distances from point to poles be \( x \) and \( 120 – x \) feet.

For first pole (60°):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3} \]

For second pole (30°):

\[ \tan 30^\circ = \frac{h}{120 – x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{120 – x} \]

Substitute \( h = x\sqrt{3} \):

\[ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{120 – x} \]

\[ 120 – x = 3x \]

\[ 4x = 120 \Rightarrow x = 30 \text{ feet} \]

Then \( h = 30\sqrt{3} \) feet

Distances: 30 feet and 90 feet

Answer: The poles are \( 30\sqrt{3} \) feet high. The point is 30 feet from one pole and 90 feet from the other.

Solution:

Let height of tower be \( h \) meters.

Let angles be \( \theta \) and \( 90^\circ – \theta \).

From first point (4m):

\[ \tan \theta = \frac{h}{4} \]

From second point (9m):

\[ \tan(90^\circ – \theta) = \frac{h}{9} \Rightarrow \cot \theta = \frac{h}{9} \]

Since \( \tan \theta \times \cot \theta = 1 \):

\[ \frac{h}{4} \times \frac{h}{9} = 1 \]

\[ h^2 = 36 \Rightarrow h = 6 \text{ meters} \]

Answer: The height of the tower is 6 meters.

Solution:

Let initial distance be \( x \) meters and height \( h = 1500\sqrt{3} \) meters.

Initial position (60°):

\[ \tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500 \text{ meters} \]

After 15 seconds (30°):

\[ \tan 30^\circ = \frac{h}{x + d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1500\sqrt{3}}{1500 + d} \]

\[ 1500 + d = 1500 \times 3 = 4500 \]

\[ d = 3000 \text{ meters} \]

Speed = Distance/Time = \( \frac{3000}{15} = 200 \) m/s

Convert to km/h: \( 200 \times \frac{18}{5} = 720 \) km/h

Answer: The speed of the jet plane is 720 km/h.

Solution:

Let height of tower be \( h_t \) and building be \( h_b \), distance between them be \( d \).

From building’s foot to tower’s top:

\[ \tan 30^\circ = \frac{h_t}{d} \Rightarrow d = h_t\sqrt{3} \]

From tower’s foot to building’s top:

\[ \tan 60^\circ = \frac{h_b}{d} \Rightarrow \sqrt{3} = \frac{h_b}{h_t\sqrt{3}} \]

\[ h_b = 3h_t \]

Ratio \( \frac{h_t}{h_b} = \frac{h_t}{3h_t} = \frac{1}{3} \)

Answer: The ratio of heights of tower to building is 1:3.