Trigonometry in right-angled triangles – Year 9 maths test answers to download as PDF

← Back to the test

1. Answer to Question 1 (4 marks)

   1. Label the sides from angle A: the 5 cm side is opposite, the 12 cm side is adjacent and the 13 cm side is the hypotenuse. Then \(\sin A=\dfrac{\text{opp}}{\text{hyp}}=\)\(\dfrac{5}{13}\) (1 mark), \(\cos A=\dfrac{\text{adj}}{\text{hyp}}=\)\(\dfrac{12}{13}\) (1 mark) and \(\tan A=\dfrac{\text{opp}}{\text{adj}}=\)\(\dfrac{5}{12}\) (1 mark).
   2. Use the inverse tangent: \(A=\tan^{-1}\left(\dfrac{5}{12}\right)=22.61\ldots\), so \(22.6°\). (1 mark)

   The inverse sine or cosine give the same angle: \(\sin^{-1}\left(\dfrac{5}{13}\right)=22.6°\). The other acute angle is \(90-22.6=67.4°\). Make sure the calculator is in degree mode.

2. Answer to Question 2 (4 marks)

   1. The hypotenuse is known and the opposite side is wanted, so use sine: \(\sin35°=\dfrac{\text{opp}}{12}\), which gives opp \(=12\sin35°\) (1 mark) \(=6.882\ldots\), so 6.88 cm (1 mark).
   2. The hypotenuse is known and the adjacent side is wanted, so use cosine: adj \(=12\cos35°\) (1 mark) \(=9.829\ldots\), so 9.83 cm (1 mark).

   Check with Pythagoras: \(6.88^{2}+9.83^{2}=47.3+96.6=143.9\), very close to \(12^{2}=144\); the small difference comes from rounding. SOH CAH TOA helps choose the ratio: write down which two sides are involved, then pick the ratio that uses them.

3. Answer to Question 3 (4 marks)

   1. The opposite and adjacent sides are known, so use tangent: \(\tan\theta=\dfrac{7}{9}\) (1 mark). Then \(\theta=\tan^{-1}\left(\dfrac{7}{9}\right)=37.87\ldots\), which is \(37.9°\) (1 mark).
   2. The adjacent side and the hypotenuse are known, so use cosine: \(\cos\theta=\dfrac{11}{15}\) (1 mark). Then \(\theta=\cos^{-1}\left(\dfrac{11}{15}\right)=42.83\ldots\), which is \(42.8°\) (1 mark).

   Sense check for (a): the opposite side is shorter than the adjacent side, so θ must be less than 45°, and 37.9° fits. In (b) the ratio 11/15 is about 0.73; since \(\cos45°\approx0.71\), the angle should be close to 45°, which it is.

4. Answer to Question 4 (4 marks)

   1. The ladder is the hypotenuse, 6 m, and the height up the wall is opposite the 70° angle at the ground. Use sine: height \(=6\sin70°\) (1 mark) \(=5.638\ldots\), so 5.64 m (1 mark).
   2. The distance from the wall is adjacent to the 70° angle. Use cosine: distance \(=6\cos70°\) (1 mark) \(=2.052\ldots\), so 2.05 m (1 mark).

   Check with Pythagoras: \(5.64^{2}+2.05^{2}=31.8+4.2=36.0=6^{2}\). A steep angle like 70° should give a tall reach and a short distance from the wall, which agrees with these answers. Safety guides recommend an angle of about 75° for ladders.

5. Answer to Question 5 (4 marks)

   1. From Sam's eyes, the horizontal distance of 45 m is adjacent to the 28° angle and the height above eye level is opposite. Use tangent: height \(=45\tan28°\) (1 mark) \(=23.93\ldots\), so 23.9 m (1 mark).
   2. Sam measured from eye level, so add his eye height: \(23.9+1.6=\)25.5 m above the ground. (1 mark)
   3. Now the opposite side is 30 m and the adjacent side is 45 m: \(\tan^{-1}\left(\dfrac{30}{45}\right)=33.69\ldots\), so \(33.7°\). (1 mark)

   Forgetting the 1.6 m in (b) is the usual error. A taller tower gives a larger angle of elevation from the same distance.

Total : 20 marks