Year 10 Trigonometry Worksheets & Practice Tests

Worked examples

Year 10 trigonometry: 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 · Angle of depression

   A lifeguard sits in a tower 8 m above the ground and looks down at a swimmer on the beach. The swimmer is 15 m from the base of the tower along flat ground. Find the angle of depression from the lifeguard to the swimmer, correct to one decimal place.

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   Key idea: The angle of depression equals the angle inside the right-angled triangle at the top (alternate angles with the horizontal). Use the tangent ratio because you have the opposite side (8 m) and the adjacent side (15 m).

   1. Label the triangle: opposite $= 8$ m, adjacent $= 15$ m.
   2. Write the trig equation:
      $\tan(\theta) = \frac{8}{15}$
   3. Solve for $\theta$:
      $\theta = \tan^{-1}\!\left(\frac{8}{15}\right)$
   4. Calculate:
      $\theta \approx 28.1°$

   Answer: The angle of depression is approximately $28.1°$.

2. Example 2 · Three-figure bearings

   A ship sails 40 km due east, then turns and sails 30 km due north. Find the bearing of the ship from its starting point, correct to the nearest degree.

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   Key idea: Draw the right-angled triangle formed by the two legs of the journey. The bearing is measured clockwise from north at the starting point, so find the angle the direct path makes with north, then convert to a three-figure bearing.

   1. Identify the right-angled triangle: the eastward leg is 40 km (opposite the required angle from north) and the northward leg is 30 km (adjacent to the required angle from north).
   2. Find the angle east of north:
      $\tan(\theta) = \frac{40}{30}$
   3. Solve:
      $\theta = \tan^{-1}\!\left(\frac{40}{30}\right) \approx 53°$
   4. Convert to a three-figure bearing (clockwise from north):
      $\text{Bearing} = 053°$

   Answer: The bearing of the ship from its starting point is $053°$.

3. Example 3 · 3D trigonometry in a rectangular prism

   A rectangular room is 6 m long, 4 m wide and 3 m high. A cable is to be run in a straight line from one bottom corner of the room to the opposite top corner. Find the length of the cable and the angle it makes with the floor, correct to one decimal place.

   Show worked solution

   Key idea: First use Pythagoras' theorem to find the diagonal of the floor, then use that diagonal and the height to find the length of the cable and the angle with the floor.

   1. Find the floor diagonal $d$:
      $d = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52}$
   2. Find the cable length $L$ using the height of 3 m:
      $L = \sqrt{(\sqrt{52})^2 + 3^2} = \sqrt{52 + 9} = \sqrt{61} \approx 7.8 \text{ m}$
   3. Find the angle $\theta$ the cable makes with the floor (opposite $= 3$ m, adjacent $= \sqrt{52}$ m):
      $\tan(\theta) = \frac{3}{\sqrt{52}}$
   4. Solve:
      $\theta = \tan^{-1}\!\left(\frac{3}{\sqrt{52}}\right) \approx 22.6°$

   Answer: The cable is approximately $7.8$ m long and makes an angle of approximately $22.6°$ with the floor.

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