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Tower and building contexts

Tower and building contexts are word problems in which heights, distances, and angles are related using trigonometry around tall structures.

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Student-friendly explanation

Towers and buildings are the most common objects in this chapter because they naturally create right triangles with the ground. The vertical side is the height of the structure, the horizontal side is the distance from the observer, and the slant side is the line of sight. Students should read the wording carefully to decide whether the question gives the height, the distance, or both. Then choose the proper ratio and solve step by step.

How to write this in exams

  1. 1

    Start with the exact idea

    Tower and building contexts are word problems in which heights, distances, and angles are related using trigonometry around tall structures.

  2. 2

    Then show how to use it

    1. Draw the tower and observer. 2. Mark the ground distance. 3. Mark the angle. 4. Choose tan, sin, or cos based on given sides. 5. Solve and check units. 6. Mention whether height is from ground or eye level.

  3. 3

    Add one concrete example

    A building is seen from a point 25 m away at 37 degrees. The height above eye level can be found using tan 37 and then adjusted if eye level is given.

  4. 4

    Avoid this incomplete answer

    A common wrong answer is to take the slant distance as the tower height. The slant is longer than the vertical height.

Definition

Tower and building contexts are word problems in which heights, distances, and angles are related using trigonometry around tall structures.

Example

A building is seen from a point 25 m away at 37 degrees. The height above eye level can be found using tan 37 and then adjusted if eye level is given.

Rule to remember

Use tan theta = height / distance for standard level-ground tower and building sketches; adjust for eye level if given.

Memory hook

Tower height stands up, ground distance lies flat.

Examples and method

Worked example

A building is 20 m away from an observer, and the angle of elevation is 45 degrees. Then tan 45 = height/20, so the height above the observer's eye is 20 m.

Method to apply

1. Draw the tower and observer. 2. Mark the ground distance. 3. Mark the angle. 4. Choose tan, sin, or cos based on given sides. 5. Solve and check units. 6. Mention whether height is from ground or eye level.

Diagram support

Sketch the tower vertically, the observer on the ground, the horizontal ground distance, and the line of sight to the top.

How CBSE asks it

Find the height of a tower or building, the distance from it, or the angle of elevation from a given distance.

Avoid common mistakes

Common confusion

Students sometimes mix the building height with the distance from the observer, or forget the eye-level correction.

Common wrong answer

A common wrong answer is to take the slant distance as the tower height. The slant is longer than the vertical height.

Exam tip

Write the height, ground distance, and angle separately on your sketch before doing any calculation.

Quick check

In a tower problem, which quantity is usually the vertical side of the triangle?

The vertical side is usually the height of the tower or building. The ground distance is the horizontal side, and the line of sight is the slant side.

Answer writing and exam use

1-mark answer

Tower and building contexts are word problems in which heights, distances, and angles are related using trigonometry around tall structures.

2-mark answer

Tower and building contexts are word problems in which heights, distances, and angles are related using trigonometry around tall structures. Use tan theta = height / distance for standard level-ground tower and building sketches; adjust for eye level if given. A building is seen from a point 25 m away at 37 degrees. The height above eye level can be found using tan 37 and then adjusted if eye level is given.

3-mark answer

Towers and buildings are the most common objects in this chapter because they naturally create right triangles with the ground. The vertical side is the height of the structure, the horizontal side is the distance from the observer, and the slant side is the line of sight. Students should read the wording carefully to decide whether the question gives the height, the distance, or both. Then choose the proper ratio and solve step by step. Use tan theta = height / distance for standard level-ground tower and building sketches; adjust for eye level if given. A building is 20 m away from an observer, and the angle of elevation is 45 degrees. Then tan 45 = height/20, so the height above the observer's eye is 20 m. Find the height of a tower or building, the distance from it, or the angle of elevation from a given distance. A common wrong answer is to take the slant distance as the tower height. The slant is longer than the vertical height.
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