Math Problem Statement

A man is 250m away from the base of the pole of height 'h' & making an angle of elevation Ø, after travelling 150m towards the pole the angle of elevation becomes 2Ø. Find the height of the pole.

Solution

To solve the problem, we will use trigonometry. Let the height of the pole be hh meters, and the angle of elevation be θ\theta (given as θ\theta and 2θ2\theta at two different distances).

Step-by-Step Solution:

  1. Initial condition: When the man is 250 meters away from the base of the pole, the angle of elevation is θ\theta.

    • Using the tangent function: tan(θ)=h250\tan(\theta) = \frac{h}{250} Therefore, we can write the equation:
    \tag{1}$$
  2. After travelling 150 meters: The man is now 250150=100250 - 150 = 100 meters away from the pole, and the angle of elevation becomes 2θ2\theta.

    • Using the tangent function again: tan(2θ)=h100\tan(2\theta) = \frac{h}{100} Therefore, we can write:
    \tag{2}$$
  3. Relationship between tan(2θ)\tan(2\theta) and tan(θ)\tan(\theta): Using the double-angle identity for tangent: tan(2θ)=2tan(θ)1tan2(θ)\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}

  4. Substitute this into equation (2):

    \tag{3}$$
  5. Equating equations (1) and (3): Since both expressions represent hh, we set them equal to each other: 250tan(θ)=1002tan(θ)1tan2(θ)250 \cdot \tan(\theta) = 100 \cdot \frac{2\tan(\theta)}{1 - \tan^2(\theta)}

  6. Simplify the equation: Cancel out tan(θ)\tan(\theta) (assuming tan(θ)0\tan(\theta) \neq 0): 250=10021tan2(θ)250 = 100 \cdot \frac{2}{1 - \tan^2(\theta)}

  7. Solve for tan2(θ)\tan^2(\theta): 250100=21tan2(θ)\frac{250}{100} = \frac{2}{1 - \tan^2(\theta)} 2.5=21tan2(θ)2.5 = \frac{2}{1 - \tan^2(\theta)} Multiply both sides by 1tan2(θ)1 - \tan^2(\theta): 2.5(1tan2(θ))=22.5(1 - \tan^2(\theta)) = 2 Expand and solve for tan2(θ)\tan^2(\theta): 2.52.5tan2(θ)=22.5 - 2.5\tan^2(\theta) = 2 2.52=2.5tan2(θ)2.5 - 2 = 2.5\tan^2(\theta) 0.5=2.5tan2(θ)0.5 = 2.5\tan^2(\theta) tan2(θ)=0.52.5=0.2\tan^2(\theta) = \frac{0.5}{2.5} = 0.2 tan(θ)=0.2\tan(\theta) = \sqrt{0.2} tan(θ)=25\tan(\theta) = \frac{\sqrt{2}}{5}

  8. Find the height hh: Now, substitute tan(θ)=25\tan(\theta) = \frac{\sqrt{2}}{5} into equation (1): h=25025=502metersh = 250 \cdot \frac{\sqrt{2}}{5} = 50\sqrt{2} \, \text{meters}

Final Answer:

The height of the pole is 50270.7150\sqrt{2} \approx 70.71 meters.

Would you like more details or have any questions?

Here are five related questions you could explore:

  1. What if the man starts 300 meters away from the pole?
  2. How would the height change if the angle of elevation doubles again after travelling 100 meters more?
  3. How is the double-angle formula for tangent derived?
  4. What is the significance of the tangent function in trigonometry?
  5. How does the angle of elevation affect the height of objects in real-life scenarios?

Tip: Always verify the conditions for using trigonometric identities, especially the restrictions on angles and domains.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Angle of Elevation
Double-Angle Identities

Formulas

tan(θ) = h/distance
tan(2θ) = 2tan(θ) / (1 - tan^2(θ))
Height of pole: h = 50√2

Theorems

Tangent Function
Double-Angle Identity for Tangent

Suitable Grade Level

Grades 10-12