Math Problem Statement

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

Solution

To find the value of pp, we need to analyze the setup of the two towers and the given angles of elevation.

Let:

  • h1=30mh_1 = 30 \, \text{m}, the height of the shorter tower.
  • h2h_2, the height of the taller tower.
  • dd, 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 6060^\circ.
  2. The angle of elevation of the top of the shorter tower from the foot of the taller tower is 3030^\circ.
  3. The distance dd is pp times the height of the shorter tower, i.e., d=ph1=30pd = p \cdot h_1 = 30p.

We need to find pp by using trigonometric relationships in these two cases.

Step 1: Set up equations using trigonometric ratios

First angle of elevation (6060^\circ)

From the foot of the shorter tower (height h1=30mh_1 = 30 \, \text{m}), looking at the top of the taller tower with an angle of elevation of 6060^\circ:

tan(60)=h2h1d\tan(60^\circ) = \frac{h_2 - h_1}{d}

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

3=h230d\sqrt{3} = \frac{h_2 - 30}{d}

Rearrange to get:

h230=3dh_2 - 30 = \sqrt{3} \cdot d

So,

h2=30+3dh_2 = 30 + \sqrt{3} \cdot d

Second angle of elevation (3030^\circ)

From the foot of the taller tower, looking at the top of the shorter tower with an angle of elevation of 3030^\circ:

tan(30)=h1d\tan(30^\circ) = \frac{h_1}{d}

Since tan(30)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}, we have:

13=30d\frac{1}{\sqrt{3}} = \frac{30}{d}

Solving for dd:

d=303d = 30 \cdot \sqrt{3}

Step 2: Find pp

Since d=ph1=30pd = p \cdot h_1 = 30p, we substitute d=303d = 30 \sqrt{3}:

303=30p30 \sqrt{3} = 30p

Divide both sides by 30:

p=3p = \sqrt{3}

Final Answer:

The value of pp is 3\sqrt{3}.

Answer: 3m\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.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Angles of Elevation
Right Triangles

Formulas

tan(θ) = opposite / adjacent
Distance formula using trigonometric ratios

Theorems

Basic Trigonometric Ratios
Pythagorean Theorem

Suitable Grade Level

Grades 10-12