Math Problem Statement

解题步骤和答案

Solution

Let's analyze and solve the problems step-by-step:

Given Information:

  1. Height of the person: 1.5 meters.
  2. Route:
    • P1 to P2 (horizontal walk).
    • P2 to P3 (uphill walk).

Problem Breakdown:

(1) Determine the actual coordinates of P2.

Given that xx represents the horizontal distance from L, and yy represents the shadow length in meters.

  • From P1 to P2, it is a horizontal walk, hence y=ay = a.
  • At P2, we have coordinates (6,a)(6, a).

(2) Find the height of the lamp LL.

  • From the first graph, the height of the lamp LL can be calculated using the similar triangles principle.
  • The height of the shadow at P2P2 is equal to aa.

(3) Given P2P3P2P3 forms an angle α\alpha with the horizontal, find α\alpha when it is θ\theta.

  • The angle of 4545^\circ forms with the horizontal distance and height.

(4) Calculate the slope of OAOA and BCBC.

  • Slope of OAOA: Using the coordinates of A(6,a)A(6, a), the slope will be a060=a6\frac{a - 0}{6 - 0} = \frac{a}{6}.
  • Slope of BCBC: Using the coordinates of B(8,b)B(8, b) and C(14,c)C(14, c), find the difference in coordinates.

Solutions:

  1. Coordinates of P2P2: The actual coordinates can be interpreted as P2(6,a)P2(6, a).

  2. Height of Lamp LL: Using similar triangles for shadow: hd=1.5y\frac{h}{d} = \frac{1.5}{y} where hh is the height of the lamp and dd is the distance from P1P1 to the person.

  3. Angle Calculation: If α=45\alpha = 45^\circ, then: tan(45)=height changehorizontal distance\tan(45^\circ) = \frac{\text{height change}}{\text{horizontal distance}} and use the given distances to calculate.

  4. Slope Calculation: Slope of OA=a6\text{Slope of } OA = \frac{a}{6} Slope of BC=cb148\text{Slope of } BC = \frac{c - b}{14 - 8}

Let's compute the height of the lamp, the coordinates, and slopes.

Detailed Steps:

1. Coordinates of P2P2:

Given coordinates P2(6,a)P2(6, a).

2. Height of Lamp LL:

Using the shadow principle: L6=1.5a    L=6×1.5a\frac{L}{6} = \frac{1.5}{a} \implies L = 6 \times \frac{1.5}{a}

3. Angle Calculation:

For α=45\alpha = 45^\circ, tan(45)=1\tan(45^\circ) = 1: 1=ΔyΔx1 = \frac{\Delta y}{\Delta x}

4. Slopes:

Slope of OAOA: a6\frac{a}{6}

Slope of BCBC: cb148=cb6\frac{c - b}{14 - 8} = \frac{c - b}{6}

If we assume a=3a = 3, b=4b = 4, and c=7c = 7, the calculations will be specific.

Would you like to see the calculations for specific values or have any questions? Here are five questions for further inquiry:

  1. Can you provide the specific values of aa, bb, and cc for precise calculation?
  2. How should we interpret the distances for the shadow and height in different scenarios?
  3. Do you need a step-by-step calculation for finding the height of the lamp?
  4. Would you like to know more about how to derive the slope using coordinates?
  5. Do you need further clarification on similar triangles and shadow calculations?

Tip:

When solving geometry problems involving shadows and heights, always look for similar triangles to simplify the calculations.

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Math Problem Analysis

Mathematical Concepts

Geometry
Similar triangles
Trigonometry
Coordinate geometry

Formulas

Distance formula
Shadow principle formula: L = 6 * (1.5 / a)
Angle calculation using tangent: tan(45°) = 1

Theorems

Similar triangles principle

Suitable Grade Level

High School