Q 42. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 600 and the angle of elevation of the top of the second tower from the
foot of the first tower is 300. The distance between the two towers is 'p' times the height of the shorter tower. What is 'p' equal to?
root 2 m 
1/2 m
root 3/2 
1/3 m
1/ root 3 m 

1/root 2 m 

root 6 m 

root 3 m

To find the value of \( p \), we need to analyze the setup of the two towers and the given angles of elevation.

Let:
- \( h_1 = 30 \, \text{m} \), the height of the shorter tower.
- \( h_2 \), the height of the taller tower.
- \( d \), the distance between the two towers.

From the problem:
1. The angle of elevation of the top of the taller tower from the foot of the shorter tower is \( 60^\circ \).
2. The angle of elevation of the top of the shorter tower from the foot of the taller tower is \( 30^\circ \).
3. The distance \( d \) is \( p \) times the height of the shorter tower, i.e., \( d = p \cdot h_1 = 30p \).

We need to find \( p \) by using trigonometric relationships in these two cases.

### Step 1: Set up equations using trigonometric ratios

#### First angle of elevation (\( 60^\circ \))
From the foot of the shorter tower (height \( h_1 = 30 \, \text{m} \)), looking at the top of the taller tower with an angle of elevation of \( 60^\circ \):

\[
\tan(60^\circ) = \frac{h_2 - h_1}{d}
\]

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

\[
\sqrt{3} = \frac{h_2 - 30}{d}
\]

Rearrange to get:

\[
h_2 - 30 = \sqrt{3} \cdot d
\]

So,

\[
h_2 = 30 + \sqrt{3} \cdot d
\]

#### Second angle of elevation (\( 30^\circ \))
From the foot of the taller tower, looking at the top of the shorter tower with an angle of elevation of \( 30^\circ \):

\[
\tan(30^\circ) = \frac{h_1}{d}
\]

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

\[
\frac{1}{\sqrt{3}} = \frac{30}{d}
\]

Solving for \( d \):

\[
d = 30 \cdot \sqrt{3}
\]

### Step 2: Find \( p \)

Since \( d = p \cdot h_1 = 30p \), we substitute \( d = 30 \sqrt{3} \):

\[
30 \sqrt{3} = 30p
\]

Divide both sides by 30:

\[
p = \sqrt{3}
\]

### Final Answer:
The value of \( p \) is \( \sqrt{3} \).

**Answer:** \(\boxed{\sqrt{3} \, \text{m}}\)

---

Do you want details or have any questions? 

Here are five related questions:

1. How can trigonometric functions be used to calculate distances in different scenarios?
2. How does the angle of elevation affect the calculation of heights?
3. What would change if the angle of elevation were different in this problem?
4. How are similar triangles useful in problems involving angles of elevation?
5. Can this method be applied to other real-world problems, like estimating mountain heights?

**Tip:** Remember that angles of elevation and depression can help solve many real-world distance and height problems using trigonometry.

The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. The distance between the two towers is 'p' times the height of the shorter tower. What is 'p' equal to?

Explore how to calculate the distance between two towers using angles of elevation and trigonometric principles. This problem involves trigonometric ratios and right triangle properties, helping students understand practical applications of angles of elevation and basic trigonometric relationships. Suitable for students in Grades 10-12.

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