Year 10A Pythagoras and Trigonometry in 3D Worksheets & Practice

Worked examples

Year 10A pythagoras and trigonometry in 3d: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Space diagonal of a rectangular prism

   A rectangular prism has length $12$ cm, width $5$ cm and height $4$ cm. Find the length of the space diagonal (the diagonal from one corner to the opposite corner through the inside of the prism). Give your answer correct to two decimal places.

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   Key idea: The space diagonal $d$ can be found using the 3D extension of Pythagoras' theorem: $d = \sqrt{l^2 + w^2 + h^2}$.

   1. Write the formula and substitute the dimensions:
      $d = \sqrt{12^2 + 5^2 + 4^2}$
   2. Evaluate each square:
      $d = \sqrt{144 + 25 + 16}$
   3. Add and take the square root:
      $d = \sqrt{185} \approx 13.60 \text{ cm}$

   Answer: The space diagonal is approximately $13.60$ cm.

2. Example 2 · Angle between a space diagonal and the base

   A rectangular prism has length $8$ m, width $6$ m and height $5$ m. Find the angle that the space diagonal makes with the base of the prism, correct to one decimal place.

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   Key idea: First find the diagonal of the base, then use trigonometry in the vertical right-angled triangle formed by that base diagonal and the space diagonal.

   1. Find the base diagonal $b$ using Pythagoras' theorem:
      $b = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ m}$
   2. The space diagonal, the base diagonal and the height form a right-angled triangle. The side opposite the required angle $\theta$ is the height ($5$ m) and the side adjacent is the base diagonal ($10$ m).
   3. Apply the tangent ratio:
      $\tan\theta = \frac{5}{10} = 0.5$
   4. Solve for $\theta$:
      $\theta = \tan^{-1}(0.5) \approx 26.6°$

   Answer: The space diagonal makes an angle of approximately $26.6°$ with the base.

3. Example 3 · Angle of elevation to a pyramid apex

   A square pyramid has a square base with side length $10$ m. Its apex is directly above the centre of the base at a height of $12$ m. A visitor stands at the midpoint of one side of the base. Find the angle of elevation from the visitor to the apex of the pyramid, correct to one decimal place.

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   Key idea: Identify the right-angled triangle formed by the horizontal distance from the visitor to the point directly below the apex, the vertical height, and the slant line of sight to the apex.

   1. The visitor stands at the midpoint of one side of the square base. The centre of the base is directly below the apex. The horizontal distance from the midpoint of a side to the centre of a square with side length $10$ m is half the side length:
      $d = \frac{10}{2} = 5 \text{ m}$
   2. The vertical height to the apex is $12$ m. These two lengths form the two shorter sides of a right-angled triangle, with the line of sight to the apex as the hypotenuse.
   3. Apply the tangent ratio, where the opposite side is the height and the adjacent side is the horizontal distance:
      $\tan\theta = \frac{12}{5} = 2.4$
   4. Solve for $\theta$:
      $\theta = \tan^{-1}(2.4) \approx 67.4°$

   Answer: The angle of elevation from the visitor to the apex is approximately $67.4°$.

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