Height And Distance

Answer: Option C
Solution:
To find the distance between the two ships, we can use trigonometry.

Given:
– Height of the lighthouse = 100 meters
– Angle of elevation from ship 1 = 30°
– Angle of elevation from ship 2 = 45°
– We need to find the distance between the two ships.

Step 1: Label the variables
Let the distance from the first ship (with 30° angle) to the base of the lighthouse be \( x_1 \) meters and the distance from the second ship (with 45° angle) to the base of the lighthouse be \( x_2 \) meters.

Step 2: Use trigonometry to find \( x_1 \) and \( x_2 \)
We can use the tangent of the angles of elevation, which is given by:

\[
\tan(\theta) = \frac{\text{height}}{\text{distance}}
\]

1. For the first ship (angle of elevation = 30°):
\[
\tan(30^\circ) = \frac{100}{x_1}
\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[
\frac{1}{\sqrt{3}} = \frac{100}{x_1}
\] Solving for \( x_1 \):
\[
x_1 = 100 \sqrt{3} \approx 100 \times 1.732 \approx 173.2 \, \text{meters}
\]

2. For the second ship (angle of elevation = 45°):
\[
\tan(45^\circ) = \frac{100}{x_2}
\] Since \( \tan(45^\circ) = 1 \), we have:
\[
1 = \frac{100}{x_2}
\] Solving for \( x_2 \):
\[
x_2 = 100 \, \text{meters}
\]

Step 3: Calculate the total distance between the ships
The total distance between the two ships is the sum of \( x_1 \) and \( x_2 \):
\[
\text{Total distance} = x_1 + x_2 = 173.2 + 100 = 273.2 \, \text{meters}
\]

Final Answer:
The distance between the two ships is approximately 273.2 meters.

Answer: Option D
Solution:
Let’s break this down step by step using trigonometry.

Given:
– Angle of elevation from point P = 30º
– Angle of elevation after the man walks closer to the tower = 60º
– Let the height of the tower be \( h \).
– The distance between the base of the tower and the point P initially is \( x \).
– After walking some distance, the man is at a new point, say point Q, where the distance from the tower is \( y \).

We need to find the initial distance \( x \) between the base of the tower and point P.

Step 1: Set up the equations using trigonometry

The tangent of an angle in a right triangle is the ratio of the opposite side (height of the tower) to the adjacent side (horizontal distance from the observer to the tower).

From point P (initial position):
\[
\tan(30^\circ) = \frac{h}{x}
\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[
\frac{1}{\sqrt{3}} = \frac{h}{x}
\] Solving for \( h \), we get:
\[
h = \frac{x}{\sqrt{3}} \quad \text{(Equation 1)}
\]

From point Q (after walking closer):
The angle of elevation becomes 60°. The distance from the tower is now \( y \). Using the tangent function again:
\[
\tan(60^\circ) = \frac{h}{y}
\] Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[
\sqrt{3} = \frac{h}{y}
\] Solving for \( h \), we get:
\[
h = \sqrt{3} \cdot y \quad \text{(Equation 2)}
\]

Step 2: Relate the distances
The man walked some distance towards the tower, so the initial distance \( x \) is the sum of the new distance \( y \) and the distance the man walked:
\[
x = y + d
\] Where \( d \) is the distance the man walked.

Step 3: Solve the system of equations
Now, we have two equations for \( h \):
1. \( h = \frac{x}{\sqrt{3}} \)
2. \( h = \sqrt{3} \cdot y \)

Equating these two expressions for \( h \):
\[
\frac{x}{\sqrt{3}} = \sqrt{3} \cdot y
\]

Multiplying both sides by \( \sqrt{3} \) to simplify:
\[
x = 3y
\]

Step 4: Final result
From the equation \( x = 3y \), we conclude that the initial distance from point P to the base of the tower is three times the distance from point Q to the base of the tower.

Final Answer:
The distance between the base of the tower and point P is 3y, where \( y \) is the distance the man walked closer to the tower.

Answer: Option D
Solution:
We can solve this problem using trigonometry.

Given:
– Angle of elevation of the ladder = 60º
– Distance of the foot of the ladder from the wall = 4.6 m
– We need to find the length of the ladder, which we’ll call \( L \).

Step 1: Use trigonometry

The situation forms a right triangle, where:
– The length of the ladder is the hypotenuse (\( L \)).
– The distance from the wall to the foot of the ladder is the base of the triangle (4.6 m).
– The height of the ladder from the ground to the top of the wall is the opposite side (which we don’t know).

We can use the cosine function to relate the angle of elevation and the adjacent side (the base of the triangle) to the hypotenuse (the ladder’s length):

\[
\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}
\]

For this problem:
\[
\cos(60^\circ) = \frac{4.6}{L}
\]

Since \( \cos(60^\circ) = \frac{1}{2} \), the equation becomes:
\[
\frac{1}{2} = \frac{4.6}{L}
\]

Step 2: Solve for \( L \)

Now, solve for \( L \):
\[
L = 4.6 \times 2 = 9.2 \, \text{meters}
\]

Final Answer:
The length of the ladder is 9.2 meters.

Answer: Option A
Solution:
We can solve this problem using trigonometry.

Given:
– The height of the observer = 1.6 m
– The distance from the observer to the tower = \( 20\sqrt{3} \) meters
– The angle of elevation from the observer’s eye to the top of the tower = 30º
– We need to find the height of the tower.

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

Step 1: Use trigonometry

The situation forms a right triangle where:
– The distance from the observer to the tower is the base of the triangle (\( 20\sqrt{3} \)).
– The vertical difference in height between the observer’s eye level and the top of the tower is the opposite side to the angle of elevation.
– The angle of elevation is 30º.

The tangent of the angle of elevation is given by the ratio of the opposite side (height difference) to the adjacent side (distance between observer and tower):

\[
\tan(\theta) = \frac{\text{height difference}}{\text{distance}}
\]

For this problem:
\[
\tan(30^\circ) = \frac{h – 1.6}{20\sqrt{3}}
\]

Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), the equation becomes:
\[
\frac{1}{\sqrt{3}} = \frac{h – 1.6}{20\sqrt{3}}
\]

Step 2: Solve for \( h \)

Now, cross-multiply to solve for \( h \):

\[
1 \times 20\sqrt{3} = (h – 1.6) \times \sqrt{3}
\]

Simplify the equation:

\[
20 = h – 1.6
\]

Now, solve for \( h \):

\[
h = 20 + 1.6 = 21.6 \, \text{meters}
\]

Final Answer:
The height of the tower is 21.6 meters.

Answer: Option C
Solution:
We can solve this problem using trigonometry.

Given:
– Angle of elevation = 30º
– Height of the tower = 100 m
– We need to find the distance of point P from the foot of the tower (let’s call it \( d \)).

Step 1: Use trigonometry

In this problem, we have a right triangle where:
– The height of the tower is the opposite side (\( 100 \) m).
– The distance from the point \( P \) to the foot of the tower is the adjacent side (\( d \)).
– The angle of elevation is \( 30^\circ \).

We can use the tangent function to relate the angle of elevation to the opposite and adjacent sides:

\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]

For this problem:
\[
\tan(30^\circ) = \frac{100}{d}
\]

Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), the equation becomes:

\[
\frac{1}{\sqrt{3}} = \frac{100}{d}
\]

Step 2: Solve for \( d \)

Now, solve for \( d \):

\[
d = 100 \times \sqrt{3}
\]

Since \( \sqrt{3} \approx 1.732 \):

\[
d = 100 \times 1.732 = 173.2 \, \text{meters}
\]

Final Answer:
The distance of point P from the foot of the tower is approximately 173.2 meters.

Answer: Option A
Solution:
We can solve this problem using trigonometry.

Given:
– The length of the shadow of the tree is \( \sqrt{3} \) times the height of the tree.
– We need to find the angle of elevation of the sun.

Let:
– The height of the tree be \( h \).
– The length of the shadow be \( \sqrt{3}h \).

Step 1: Use trigonometry

In this situation, we have a right triangle formed by:
– The height of the tree (\( h \)) as the opposite side.
– The length of the shadow (\( \sqrt{3}h \)) as the adjacent side.
– The angle of elevation of the sun, which we will call \( \theta \).

We can use the tangent function, which is the ratio of the opposite side to the adjacent side:

\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{\sqrt{3}h}
\]

Simplifying:

\[
\tan(\theta) = \frac{1}{\sqrt{3}}
\]

Step 2: Find the angle \( \theta \)

We know that:

\[
\tan(30^\circ) = \frac{1}{\sqrt{3}}
\]

Thus, the angle of elevation of the sun is:

\[
\theta = 30^\circ
\]

Final Answer:
The angle of elevation of the sun is 30º.

Answer: Option C
Solution:
Let AB = h then,
= tan =
AB
AC
30
∘ 1
3
– √
⇒ AC = AB × 3 = h And,
– √ 3
– √
⇒ = tan = 1 AB
AD
45
∘
⇒ AD = AB = h
CD = 20
⇒ (AC − AD) = 20
⇒ h 3 − h = 20
– √
∴ h = ×
20
( 3 − 1)
– √
( 3 + 1)
– √
( 3 + 1)
– √
= 10 ( 3 + 1) m
– √
= (10 × 2.73) m
= 27.3m

Answer: Option A
Solution:
To solve this, we need to use trigonometry and work through the problem step by step.

Given:
– Height of the tower = 150 m
– Angle of depression for the first object = 45°
– Angle of depression for the second object = 60°
– We need to find the distance between the two objects.

Step 1: Define the variables

Let:
– \( x_1 \) be the horizontal distance from the base of the tower to the first object (where the angle of depression is 45°).
– \( x_2 \) be the horizontal distance from the base of the tower to the second object (where the angle of depression is 60°).

Step 2: Use trigonometry

The angle of depression is the angle formed between the line of sight from the top of the tower and the horizontal. The horizontal distance and the height of the tower form right triangles.

For the first object:
– The angle of depression is 45°, so we use \( \tan(45^\circ) \).
\[
\tan(45^\circ) = \frac{\text{height of the tower}}{\text{distance from the base of the tower to the object}}
\] Since \( \tan(45^\circ) = 1 \):
\[
1 = \frac{150}{x_1}
\] Solving for \( x_1 \):
\[
x_1 = 150 \, \text{m}
\]

For the second object:
– The angle of depression is 60°, so we use \( \tan(60^\circ) \).
\[
\tan(60^\circ) = \frac{\text{height of the tower}}{\text{distance from the base of the tower to the object}}
\] Since \( \tan(60^\circ) = \sqrt{3} \):
\[
\sqrt{3} = \frac{150}{x_2}
\] Solving for \( x_2 \):
\[
x_2 = \frac{150}{\sqrt{3}} = 50\sqrt{3} \, \text{m}
\]

Step 3: Calculate the distance between the two objects

The distance between the two objects is the difference between the two distances \( x_1 \) and \( x_2 \):
\[
\text{Distance between objects} = x_1 – x_2 = 150 – 50\sqrt{3}
\]

Step 4: Approximate the result

We know that \( \sqrt{3} \approx 1.732 \), so:

\[
50\sqrt{3} \approx 50 \times 1.732 = 86.6 \, \text{m}
\]

Thus, the distance between the two objects is:

\[
150 – 86.6 = 63.4 \, \text{m}
\]

Final Answer:
The distance between the two objects is approximately 63.4 meters.

Answer: Option B
Solution:
We can solve this using trigonometry.

Given:
– Distance from the base of the tower to the point = 70 m
– Angle of depression = 60°
– We need to find the height of the tower, which we’ll denote as \( h \).

Step 1: Use trigonometry

The angle of depression is the angle between the horizontal and the line of sight from the top of the tower to the point on the ground. Since the angle of depression is given, the corresponding angle of elevation from the point to the top of the tower will also be 60° (by alternate interior angles).

Now, we can use the tangent function, which is defined as the ratio of the opposite side (height of the tower) to the adjacent side (distance from the base of the tower):

\[
\tan(60^\circ) = \frac{\text{height of the tower}}{\text{distance from the base of the tower to the point}}
\]

Since \( \tan(60^\circ) = \sqrt{3} \), we get the equation:

\[
\sqrt{3} = \frac{h}{70}
\]

Step 2: Solve for \( h \)

Multiply both sides by 70 to solve for \( h \):

\[
h = 70 \times \sqrt{3}
\]

Using \( \sqrt{3} \approx 1.732 \):

\[
h = 70 \times 1.732 \approx 121.24 \, \text{m}
\]

Final Answer:
The height of the tower is approximately 121.24 meters.

Answer: Option C
Solution:
To solve this, we can use trigonometry.

Given:
– The angle of depression changes from 30° to 45° over a period of 12 minutes.
– The car is moving towards the observation tower at a uniform speed.
– We need to find how soon after the angle of depression becomes 45° the car will reach the tower.

Step 1: Set up the situation

Let:
– \( h \) be the height of the observation tower.
– \( x_1 \) be the distance of the car from the tower when the angle of depression is 30°.
– \( x_2 \) be the distance of the car from the tower when the angle of depression is 45°.

We are given that the car moves towards the tower from a distance where the angle of depression is 30° to a distance where the angle of depression is 45°.

Step 2: Use trigonometry

For the angle of depression of 30° (at distance \( x_1 \)):
\[
\tan(30^\circ) = \frac{h}{x_1}
\] Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we get:
\[
\frac{1}{\sqrt{3}} = \frac{h}{x_1}
\] Thus:
\[
x_1 = h\sqrt{3} \quad \text{(Equation 1)}
\]

For the angle of depression of 45° (at distance \( x_2 \)):
\[
\tan(45^\circ) = \frac{h}{x_2}
\] Since \( \tan(45^\circ) = 1 \), we get:
\[
1 = \frac{h}{x_2}
\] Thus:
\[
x_2 = h \quad \text{(Equation 2)}
\]

Step 3: Find the distance the car covers

The car moves from a distance of \( x_1 \) to \( x_2 \) over 12 minutes. The distance traveled by the car is:
\[
x_1 – x_2 = h\sqrt{3} – h = h(\sqrt{3} – 1)
\]

Step 4: Find the speed of the car

The car covers this distance \( h(\sqrt{3} – 1) \) in 12 minutes. To find the speed of the car, we can write:
\[
\text{Speed of the car} = \frac{h(\sqrt{3} – 1)}{12} \quad \text{(in minutes)}
\]

Step 5: Calculate how soon the car will reach the tower

The car will reach the tower when the distance is \( x_2 = h \). The time it takes for the car to cover this remaining distance is:
\[
\text{Time to reach the tower} = \frac{h}{\text{Speed of the car}}
\] Substituting the speed of the car from the previous step:
\[
\text{Time to reach the tower} = \frac{h}{\frac{h(\sqrt{3} – 1)}{12}} = \frac{12}{\sqrt{3} – 1}
\]

Step 6: Simplify the expression

We can rationalize the denominator by multiplying both the numerator and denominator by \( \sqrt{3} + 1 \):
\[
\frac{12}{\sqrt{3} – 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{12(\sqrt{3} + 1)}{(\sqrt{3} – 1)(\sqrt{3} + 1)}
\] Using the difference of squares formula:
\[
(\sqrt{3} – 1)(\sqrt{3} + 1) = 3 – 1 = 2
\] Thus:
\[
\text{Time to reach the tower} = \frac{12(\sqrt{3} + 1)}{2} = 6(\sqrt{3} + 1)
\]

Using \( \sqrt{3} \approx 1.732 \), we get:
\[
\text{Time to reach the tower} = 6(1.732 + 1) = 6 \times 2.732 = 16.392 \, \text{minutes}
\]

Final Answer:
The car will reach the observation tower in approximately 16.39 minutes after the angle of depression becomes 45°.

Answer: Option A
Solution:
In this scenario, we have two boys: one at the top of the tower and the other at the ground. Let’s break down the key points:

1. Angle of Elevation: This is the angle formed by the line of sight from the boy at the ground (who is looking up) to the boy at the top of the tower. The angle is measured from the horizontal ground up to the line of sight.

2. Angle of Depression: This is the angle formed by the line of sight from the boy at the top of the tower (who is looking down) to the boy on the ground. The angle is measured from the horizontal line at the top of the tower down to the line of sight.

For both boys to be able to see each other:
– The angle of elevation for the boy on the ground looking up to the top of the tower is the same as the angle of depression for the boy at the top of the tower looking down, because the two lines of sight are parallel (assuming the ground is level).

So, the angle of elevation and depression are equal**. This means the two angles would have the same measure, but the actual value depends on the height of the tower and the distance between the two boys.

If you have specific values for the height of the tower or the distance, we can calculate the exact angles!

Answer: Option B
Solution:
Given that the height of the vertical pole is \(\sqrt{3}\) times the length of its shadow, we can use trigonometry to find the angle of elevation of the sun.

Let:
– The height of the pole = \( h = \sqrt{3} \times L \), where \( L \) is the length of the shadow.
– The angle of elevation of the sun is \( \theta \).

We know that the tangent of the angle of elevation is the ratio of the height of the pole to the length of the shadow:

\[
\tan(\theta) = \frac{h}{L}
\]

Substitute \( h = \sqrt{3} \times L \) into the equation:

\[
\tan(\theta) = \frac{\sqrt{3} \times L}{L} = \sqrt{3}
\]

Now, we can find the angle \( \theta \) by taking the inverse tangent (arctan) of \( \sqrt{3} \):

\[
\theta = \tan^{-1}(\sqrt{3})
\]

We know that:

\[
\tan(60^\circ) = \sqrt{3}
\]

Thus:

\[
\theta = 60^\circ
\]

Conclusion:
The angle of elevation of the sun at that time is 60°.

Answer: Option D
Solution:
Let’s break this problem into manageable steps.

Given:
– The two persons are \( a \) meters apart.
– The height of one person is double that of the other.
– The observer is at the midpoint of the line joining their feet.
– The angular elevations to their tops are complementary, meaning the angles add up to 90°.

Let’s define:
– Let the height of the shorter person be \( h \).
– Then, the height of the taller person is \( 2h \).
– The distance between the two persons is \( a \), so the observer is at a distance of \( \frac{a}{2} \) from both persons.

Step 1: Use the tangent function for angular elevation.

The tangent of the angle of elevation is the ratio of the height of the person to the distance from the observer. Let’s assume the angle of elevation to the shorter person is \( \theta_1 \) and to the taller person is \( \theta_2 \).

From the information that the angles are complementary:

\[
\theta_1 + \theta_2 = 90^\circ
\]

So, the tangent of the angles can be written as:

\[
\tan(\theta_1) = \frac{h}{\frac{a}{2}} = \frac{2h}{a}
\]

\[
\tan(\theta_2) = \frac{2h}{\frac{a}{2}} = \frac{4h}{a}
\]

Since \( \theta_1 \) and \( \theta_2 \) are complementary, we use the identity for complementary angles:

\[
\tan(\theta_1) \times \tan(\theta_2) = 1
\]

Substituting the values for \( \tan(\theta_1) \) and \( \tan(\theta_2) \):

\[
\left( \frac{2h}{a} \right) \times \left( \frac{4h}{a} \right) = 1
\]

Simplifying the equation:

\[
\frac{8h^2}{a^2} = 1
\]

\[
8h^2 = a^2
\]

\[
h^2 = \frac{a^2}{8}
\]

\[
h = \frac{a}{\sqrt{8}} = \frac{a}{2\sqrt{2}}
\]

Conclusion:
The height of the shorter person is \( \frac{a}{2\sqrt{2}} \) meters.

Answer: Option A
Solution:
To find the height of the tower, we can use trigonometry. Specifically, we’ll use the tangent of the angle of elevation.

Given:
– The angle of elevation to the top of the tower is \( 60^\circ \).
– The horizontal distance from the observer to the foot of the tower is \( 100 \) meters.
– We need to find the height of the tower \( h \).

Using the tangent function:

\[
\tan(\theta) = \frac{\text{height of the tower}}{\text{distance from the foot of the tower}}
\]

Substitute the given values:

\[
\tan(60^\circ) = \frac{h}{100}
\]

We know that \( \tan(60^\circ) = \sqrt{3} \), so:

\[
\sqrt{3} = \frac{h}{100}
\]

Now, solve for \( h \):

\[
h = 100 \times \sqrt{3}
\]

Using \( \sqrt{3} \approx 1.732 \):

\[
h = 100 \times 1.732 \approx 173.2 \, \text{meters}
\]

Conclusion:
The height of the tower is approximately 173.2 meters.

Answer: Option C
Solution:
To find the distance of the car from the base of the tower, we will use trigonometry.

Given:
– The height of the tower is 75 meters.
– The angle of depression to the car is 30°.
– We need to find the horizontal distance between the car and the base of the tower.

Using the tangent function:

The tangent of the angle of depression is the ratio of the height of the tower to the horizontal distance. Since the angle of depression is from the top of the tower to the car on the ground, this forms a right-angled triangle.

Let the horizontal distance be \( x \).

\[
\tan(30^\circ) = \frac{\text{height of the tower}}{\text{distance from the base}}
\]

Substitute the known values:

\[
\tan(30^\circ) = \frac{75}{x}
\]

We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:

\[
\frac{1}{\sqrt{3}} = \frac{75}{x}
\]

Now, solve for \( x \):

\[
x = 75 \times \sqrt{3}
\]

Using \( \sqrt{3} \approx 1.732 \):

\[
x = 75 \times 1.732 \approx 129.9 \, \text{meters}
\]

Conclusion:
The distance of the car from the base of the tower is approximately 129.9 meters.

Answer: Option B
Solution:
To find the height of the wall, we can use trigonometry, specifically the sine function, because we have the length of the ladder (hypotenuse), the angle it makes with the wall, and the height of the wall (opposite side).

Given:
– Length of the ladder (hypotenuse) = 15 meters
– Angle with the wall = 60°
– We need to find the height of the wall (opposite side).

Using the sine function:
The sine of an angle in a right-angled triangle is the ratio of the opposite side to the hypotenuse:

\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]

Substitute the known values:

\[
\sin(60^\circ) = \frac{h}{15}
\]

We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), so:

\[
\frac{\sqrt{3}}{2} = \frac{h}{15}
\]

Now, solve for \( h \):

\[
h = 15 \times \frac{\sqrt{3}}{2}
\]

Using \( \sqrt{3} \approx 1.732 \):

\[
h = 15 \times \frac{1.732}{2} \approx 15 \times 0.866 \approx 12.99 \, \text{meters}
\]

Conclusion:
The height of the wall is approximately 13 meters.

Answer: Option B
Solution:
Let’s work through this step by step to find the height of the smaller pole.

Given:
– Two poles are \( a \) meters apart.
– The height of one pole is double the height of the other.
– From the middle point of the line joining their feet, the angular elevations of their tops are complementary, i.e., the angles add up to 90°.

Let:
– The height of the smaller pole be \( h \).
– The height of the taller pole is \( 2h \).
– The distance from the observer (at the midpoint of the line joining their feet) to both poles is \( \frac{a}{2} \).

Step 1: Use trigonometry.
Let the angle of elevation to the top of the smaller pole be \( \theta_1 \), and the angle of elevation to the top of the taller pole be \( \theta_2 \).

The tangent of the angle of elevation is the ratio of the height to the horizontal distance. Thus, we have:

\[
\tan(\theta_1) = \frac{h}{\frac{a}{2}} = \frac{2h}{a}
\]

\[
\tan(\theta_2) = \frac{2h}{\frac{a}{2}} = \frac{4h}{a}
\]

Step 2: Use the complementary angle relationship.
Since the angles are complementary:

\[
\theta_1 + \theta_2 = 90^\circ
\]

Using the identity for complementary angles, we know:

\[
\tan(\theta_1) \times \tan(\theta_2) = 1
\]

Substitute the expressions for \( \tan(\theta_1) \) and \( \tan(\theta_2) \):

\[
\left( \frac{2h}{a} \right) \times \left( \frac{4h}{a} \right) = 1
\]

Simplifying:

\[
\frac{8h^2}{a^2} = 1
\]

Multiply both sides by \( a^2 \):

\[
8h^2 = a^2
\]

Solve for \( h \):

\[
h^2 = \frac{a^2}{8}
\]

\[
h = \frac{a}{\sqrt{8}} = \frac{a}{2\sqrt{2}}
\]

Conclusion:
The height of the smaller pole is \( \frac{a}{2\sqrt{2}} \) meters.

Answer: Option B
Solution:
Let’s work through the problem using trigonometry.

Given:
– The height of the cliff is 25 meters.
– The angle of elevation to the top of the tower is equal to the angle of depression to the foot of the tower.
– We need to find the height of the tower.

Let:
– The height of the tower be \( h \).
– The horizontal distance from the base of the cliff to the base of the tower be \( x \).

Step 1: Use the angles.
Let the angle of elevation to the top of the tower be \( \theta \), and since the angle of depression to the foot of the tower is the same, the angle of depression is also \( \theta \).

Step 2: Use trigonometry.
For the angle of elevation:

\[
\tan(\theta) = \frac{h – 25}{x}
\]

For the angle of depression, we have:

\[
\tan(\theta) = \frac{25}{x}
\]

Step 3: Set the two equations equal.
Since both expressions represent \( \tan(\theta) \), we can equate them:

\[
\frac{h – 25}{x} = \frac{25}{x}
\]

Step 4: Solve for \( h \).
The \( x \) terms cancel out:

\[
h – 25 = 25
\]

\[
h = 50
\]

Conclusion:
The height of the tower is 50 meters.

Answer: Option B
Solution:
To find the height of the tower, we can use trigonometry.

Given:
– The horizontal distance from the point on the ground to the foot of the tower is \( 50 \) meters.
– The angle of elevation to the top of the tower is \( 45^\circ \).

Using the tangent function:
The tangent of the angle of elevation is the ratio of the height of the tower (opposite side) to the horizontal distance (adjacent side).

\[
\tan(\theta) = \frac{\text{height of the tower}}{\text{distance from the foot of the tower}}
\]

Substitute the known values:

\[
\tan(45^\circ) = \frac{h}{50}
\]

We know that \( \tan(45^\circ) = 1 \), so:

\[
1 = \frac{h}{50}
\]

Now, solve for \( h \):

\[
h = 50 \, \text{meters}
\]

Conclusion:
The height of the tower is 50 meters.

Answer: Option A
Solution:
The ratio of the length of the rod and its shadow is given as \(1 : \sqrt{3}\).

This suggests that the tangent of the angle of elevation \(\theta\) is equal to the ratio of the height of the rod (which is the vertical side of the right triangle) to the length of its shadow (which is the horizontal side).

Thus:

\[
\tan(\theta) = \frac{\text{height of the rod}}{\text{length of the shadow}} = \frac{1}{\sqrt{3}}
\]

We know that:

\[
\tan(30^\circ) = \frac{1}{\sqrt{3}}
\]

Therefore, the angle of elevation of the sun is:

\[
\theta = 30^\circ
\]

Answer: Option C
Solution:
Let’s solve this step by step.

Given:
– The angle of elevation of the top of the chimney from the first point is \(30^\circ\).
– The angle of elevation from the second point, after walking \(x\) meters towards the chimney, changes to \(60^\circ\).
– Let the height of the chimney be \(h\).
– The distance from the first point to the base of the chimney is \(d\).

Step 1: Set up the first equation (at the first point)

From the first point, where the angle of elevation is \(30^\circ\), we can use the tangent of the angle:

\[
\tan(30^\circ) = \frac{h}{d}
\]

Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:

\[
\frac{1}{\sqrt{3}} = \frac{h}{d}
\]

So,

\[
h = \frac{d}{\sqrt{3}} \quad \text{(Equation 1)}
\]

Step 2: Set up the second equation (at the second point)

At the second point, after walking \(x\) meters towards the chimney, the angle of elevation becomes \(60^\circ\). The new distance from the base of the chimney is \(d – x\), and the angle of elevation is \(60^\circ\).

Using the tangent function again:

\[
\tan(60^\circ) = \frac{h}{d – x}
\]

Since \( \tan(60^\circ) = \sqrt{3} \), we have:

\[
\sqrt{3} = \frac{h}{d – x}
\]

So,

\[
h = \sqrt{3}(d – x) \quad \text{(Equation 2)}
\]

Step 3: Solve the system of equations

Now, we have two equations:

1. \( h = \frac{d}{\sqrt{3}} \)
2. \( h = \sqrt{3}(d – x) \)

Equating the two expressions for \(h\):

\[
\frac{d}{\sqrt{3}} = \sqrt{3}(d – x)
\]

Multiply both sides of the equation by \(\sqrt{3}\) to eliminate the fraction:

\[
d = 3(d – x)
\]

Now simplify and solve for \(d\):

\[
d = 3d – 3x
\]

\[
d – 3d = -3x
\]

\[
-2d = -3x
\]

\[
d = \frac{3x}{2}
\]

Step 4: Find the height of the chimney

Now substitute \( d = \frac{3x}{2} \) into Equation 1 to find \( h \):

\[
h = \frac{d}{\sqrt{3}} = \frac{\frac{3x}{2}}{\sqrt{3}} = \frac{3x}{2\sqrt{3}} = \frac{x\sqrt{3}}{2}
\]

Thus, the height of the chimney is:

\[
h = \frac{x\sqrt{3}}{2}
\]