Trigonometry

16 Grade 9 Mathematics questions on Trigonometry, with worked marking schemes. 16 questions are available on this topic in full.

1. Question 1

   1 mark

   cos θ = ___

   A. hypotenuse/adjacent

   B. opposite/adjacent

   C. opposite/hypotenuse

   D. adjacent/hypotenuse

   (1 mark)

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2. Question 2

   3 marks

   A ladder $5\text{m}$ long leans against a vertical wall and makes an angle of $60^{\circ }$ with the level ground. Find the height, $h$, that the ladder reaches up the wall. (Give your answer to 2 decimal places.)

   (3 marks)

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3. Question 3

   4 marks

   Given sin θ = 3/5 and θ is acute, find cos θ and tan θ without using a calculator.

   (4 marks)

   Marking scheme

   Use the identity ${\sin }^2\theta +{\cos }^2\theta =1$, which holds for every angle. (1)

   ${\cos }^2\theta =1-{\left(\frac{3}{5}\right)}^2=1-\frac{9}{25}=\frac{16}{25}$. (1)

   $\theta$ is acute, so cosine is positive: $\cos \theta =\frac{4}{5}$. (1)

   Then $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{3/5}{4/5}=\frac{3}{4}$. This is the $3,4,5$ right-angled triangle. (1)

   Final answer: $\cos \theta =\frac{4}{5}$ and $\tan \theta =\frac{3}{4}$.

4. Question 4

   3 marks

   A flagpole casts a shadow 12 m long when the angle of elevation of the sun is 30°. Find the height of the flagpole. (tan 30° = 1/√3 ≈ 0.577)

   (3 marks)

   Marking scheme

   The pole, its shadow and the sun's ray form a right-angled triangle. The height is opposite the angle of elevation and the shadow is adjacent to it, so use $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$. (1)

   $\tan 30^{\circ }=\frac{h}{12}$, so $h=12\times \tan 30^{\circ }$. (1)

   $12\times 0.577=6.924$. (1)

   Final answer: about $6.93$ m.

5. Question 5

   4 marks

   In triangle PQR, P = 90°, PR = 5 cm, and angle Q = 30°. Find QR and PQ.

   (4 marks)

   Marking scheme

   Angle $P=90^{\circ }$, so $QR$ is the hypotenuse, and $PR$ is opposite angle $Q$. That pairing calls for the sine ratio: $\sin Q=\frac{\text{opposite}}{\text{hypotenuse}}$. (1)

   $\sin 30^{\circ }=\frac{5}{QR}$, and $\sin 30^{\circ }=0.5$, so $0.5\times QR=5$ and $QR=10$ cm. (1)

   Now $PQ$ is the third side of a right-angled triangle, so use Pythagoras: $PQ=\sqrt{10^2-5^2}$. (1)

   $100-25=75$, and $\sqrt{75}=5\sqrt{3}\approx 8.66$. (1)

   Final answer: $QR=10$ cm and $PQ=5\sqrt{3}\approx 8.66$ cm.

6. Question 6

   1 mark

   tan 45° = ___

   A. $\frac{\sqrt{2}}{2}$

   B. 0

   C. √2

   D. 1

   (1 mark)

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7. Question 7

   1 mark

   sin 30° = ___

   A. $\frac{1}{2}$

   B. $\frac{\sqrt{3}}{2}$

   C. $\frac{1}{\sqrt{2}}$

   D. 1

   (1 mark)

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8. Question 8

   4 marks

   In right triangle ABC, angle B = 90°, AB = 6 cm, BC = 8 cm. Find AC, sin A, cos A, and tan A.

   (4 marks)

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9. Question 9

   1 mark

   cos 60° = ___

   A. $\frac{1}{2}$

   B. 1

   C. 0

   D. $\frac{\sqrt{3}}{2}$

   (1 mark)

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10. Question 10

    3 marks

    A boat is 50 m from the base of a cliff. The angle of elevation to the top of the cliff is 60°. Find the height of the cliff. (tan 60° = √3 ≈ 1.732)

    (3 marks)

    Marking scheme

    The cliff, the water and the line of sight form a right-angled triangle. The height is opposite the angle of elevation and the $50$ m is adjacent, so use $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$. (1)

    $\tan 60^{\circ }=\frac{h}{50}$, so $h=50\times \tan 60^{\circ }$. (1)

    $50\times 1.732=86.6$. (1)

    Final answer: $86.6$ m.

11. Question 11

    1 mark

    tan 30° = ___

    A. 1

    B. $\sqrt{3}$

    C. $\frac{1}{2}$

    D. $\frac{1}{\sqrt{3}}$

    (1 mark)

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12. Question 12

    1 mark

    In a 3-4-5 right triangle, what is sin of the angle opposite the side of length 3?

    A. $\frac{4}{5}$

    B. $\frac{3}{4}$

    C. $\frac{5}{3}$

    D. $\frac{3}{5}$

    (1 mark)

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13. Question 13

    1 mark

    A ladder 10 m long leans against a wall making an angle of 60° with the ground. How high up the wall does it reach? (sin 60° ≈ 0.866)

    A. 5 m

    B. 8.66 m

    C. 6 m

    D. 10 m

    (1 mark)

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14. Question 14

    1 mark

    In a right triangle, sin θ = ___

    A. adjacent/hypotenuse

    B. opposite/adjacent

    C. opposite/hypotenuse

    D. hypotenuse/opposite

    (1 mark)

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15. Question 15

    1 mark

    sin² θ + cos² θ = ___

    A. 2

    B. sin 2θ

    C. 0

    D. 1

    (1 mark)

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16. Question 16

    1 mark

    An angle of elevation from a point 30 m from a tree to its top is 45°. The height of the tree is ___

    A. 60 m

    B. 45 m

    C. 30 m

    D. 15 m

    (1 mark)

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