Page 2: Solutions for Problems 1 and 2

This page provides detailed solutions for the first two problems of the exam.

Problem 1 Solution: The height of the fir tree is calculated using the tangent function. Given a shadow length of 27m and a sun angle of 38.5°, the tree's height is determined to be 21.48m.

Example: tan(38.5°) × 27m = height of the tree

Problem 2 Solution: For the isosceles triangle with base c = 22m and angle a = 80°, the solution involves:

1. Calculating the remaining angles $y = 50°$
2. Using the cosine function to find the length of the equal sides (b ≈ 63.38m)

Highlight: These solutions demonstrate the practical application of Trigonometrie Textaufgaben und Lösungen Klasse 10 PDF.

Page 3: Solution for Problem 3

This page focuses on solving Problem 3, which involves a triangle with two known sides $a = 73m, b = 64m$ and one known angle $α = 80°$.

The solution process includes:

1. Using the law of sines to find angle β
2. Calculating angle γ using the fact that angles in a triangle sum to 180°
3. Applying the law of cosines to determine the length of side c

Example: c² = a² + b² - 2ab cos(γ)

The final results are:

• c ≈ 44.41m
• β ≈ 62.81°
• γ ≈ 37.19°

Highlight: This problem showcases the application of multiple trigonometric concepts in a single question, typical of Trigonometrie Aufgaben Klasse 10 Gymnasium.

Page 4: Solutions for Problems 4 and 5

This page provides solutions for Problems 4 and 5, demonstrating advanced applications of trigonometry.

Problem 4 Solution: To find the distance between the endpoints of two paths that diverge at a 98° angle with lengths 4.1km and 5.8km, the law of cosines is applied:

d² = 4.1² + 5.8² - 2(4.1)(5.8)cos(98°)

The result shows the endpoints are approximately 7.56km apart.

Problem 5 Solution: This problem involves calculating a mountain's height using two elevation angles $α = 30.11° and β = 35.25°$ measured from the ends of a 200m baseline.

The solution uses trigonometric ratios and the law of sines to determine:

1. The distance from point A to the mountain's base (1288.42m)
2. The mountain's height (646.38m)

Highlight: These problems exemplify real-world applications of trigonometry, aligning with Anwendungsaufgaben Trigonometrie.

Page 5: Continuation of Problem 5 and Start of Problem 6

This page concludes the solution for Problem 5 and begins Problem 6.

Problem 5 (continued): The page shows the final calculations confirming the mountain's height as 646.38m, demonstrating the consistency of the trigonometric approach from both measurement points.

Problem 6: This problem involves calculating the surface area of a square pyramid with base side length a = 4cm and slant height hk = 14cm.

The solution begins by calculating the pyramid's height $h$ using the Pythagorean theorem:

h² = hk² - $a/2$² h ≈ 13.66cm

Vocabulary: Slant height - The distance from the apex of a pyramid to the middle of one of the base edges.

Highlight: This problem introduces Körperberechnung Aufgaben Klasse 10 PDF, focusing on 3D geometry calculations.

Page 6: Completion of Problem 6

This page completes the solution for Problem 6, calculating the surface area of the square pyramid.

The solution involves:

1. Calculating the area of the square base: 4² = 16 cm²
2. Determining the area of one triangular face: $1/2$ × 4 × 14.14 ≈ 28.28 cm²
3. Multiplying the face area by 4 and adding the base area

Total surface area = 4 × 28.28 + 16 ≈ 129.12 cm²

Example: Surface Area = Base Area + (Number of Sides × Triangular Face Area)

Highlight: This problem demonstrates the application of both 2D and 3D geometry concepts, typical in Körperberechnung Aufgaben Klasse 10 mit Lösungen.

Page 7: Solution for Problem 7 (Part A)

This page begins the solution for Problem 7, which involves calculating the surface area of a half-cylinder water trough with a diameter of 1.30m and length of 3.85m.

Part A of the solution focuses on determining the amount of zinc sheet needed to construct the trough:

1. Calculating the area of the rectangular sides: 1.30m × 3.85m × 2 = 10.01m²
2. Determining the area of the semicircular end caps: π × $1.30/2$² ÷ 2 × 2 ≈ 1.33m²
3. Adding these areas together: 10.01m² + 1.33m² = 11.34m²

Vocabulary: Half-cylinder - A three-dimensional shape formed by cutting a cylinder in half lengthwise.

Highlight: This problem combines concepts from Körperberechnung Klasse 10 Gymnasium with practical applications.

Page 8: Solution for Problem 7 (Part B)

This final page completes the solution for Problem 7, calculating the volume of water the half-cylinder trough can hold.

The volume is calculated using the formula for a cylinder, halved:

V = (π × r² × length) ÷ 2 V = (π × (0.65m)² × 3.85m) ÷ 2 V ≈ 2.56 m³

Converting to liters: 2.56 m³ = 2,560 liters

Example: 1 m³ = 1,000 liters

Highlight: This problem showcases the practical application of volume calculations, an important aspect of Geometrie Aufgaben Klasse 10 PDF.

Page 1: Introduction and Problem Set

This page introduces a mathematics exam for 10th-grade students, presenting seven challenging problems covering trigonometry and geometry.

Key problems include:

1. Calculating the height of a fir tree using its shadow length and sun angle
2. Solving for unknown values in an isosceles triangle
3. Determining side lengths and angles in a triangle
4. Finding the distance between endpoints of two paths
5. Calculating a mountain's height using elevation angles
6. Computing the surface area of a square pyramid
7. Determining the surface area and volume of a half-cylinder water trough

Highlight: The exam covers a wide range of trigonometric and geometric concepts, emphasizing practical applications and problem-solving skills.

Vocabulary: Isosceles triangle - A triangle with two equal sides and two equal angles.