Trigonometry in Right-Angled Triangles

1. 1 Work out [5 marks]

   The diagram shows a right-angled triangle. \(\sin\theta = \dfrac{5}{13}\). (a) Work out the value of \(x\). (3 marks) (b) Work out the length of the side next to angle \(\theta\). (2 marks)

   

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   Model answer

   (a) \(\sin\theta = \dfrac{x}{39}\), so \(\dfrac{x}{39} = \dfrac{5}{13}\) and \(x = \dfrac{5}{13} \times 39 = 15\) cm. (b) By Pythagoras, \(39^2 - 15^2 = 1521 - 225 = 1296\), so the side is \(\sqrt{1296} = 36\) cm.

   Mark scheme

   • (a) \(\dfrac{x}{39} = \dfrac{5}{13}\) — M1
   • (a) \(39 \div 13 = 3\) or \(\dfrac{5 \times 39}{13}\) — M1
   • (a) 15 — A1
   • (b) \(39^2 - 15^2\) — M1
   • (b) 36 — A1
2. 2 Write down [2 marks]

   (a) Write down the exact value of \(\sin 30^\circ\). (1 mark) (b) Write down the exact value of \(\tan 45^\circ\). (1 mark)

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   Model answer

   (a) \(\sin 30^\circ = \dfrac{1}{2}\). (b) \(\tan 45^\circ = 1\).

   Mark scheme

   • (a) \(\dfrac{1}{2}\) — B1
   • (b) 1 — B1
3. 3 Work out [3 marks]

   A right-angled triangle has a hypotenuse of 14 cm and an angle of \(30^\circ\). Work out the length of the side opposite the \(30^\circ\) angle.

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   Model answer

   \(\sin 30^\circ = \dfrac{x}{14}\) and \(\sin 30^\circ = \dfrac{1}{2}\), so \(x = 7\) cm.

   Mark scheme

   • \(\sin 30^\circ = \dfrac{x}{14}\) — M1
   • \(\dfrac{1}{2} = \dfrac{x}{14}\) — M1
   • 7 cm — A1
4. 4 Work out [3 marks]

   In a right-angled triangle, the side opposite angle \(\theta\) is 7 cm and the side adjacent to angle \(\theta\) is also 7 cm. Work out the size of angle \(\theta\).

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   Model answer

   \(\tan\theta = \dfrac{7}{7} = 1\), so \(\theta = 45^\circ\).

   Mark scheme

   • \(\tan\theta = \dfrac{O}{A}\) or \(\dfrac{7}{7}\) — M1
   • \(\tan\theta = 1\) — M1
   • \(45^\circ\) — A1
5. 5 Work out [3 marks]

   Triangle \(ABC\) is right-angled at \(B\). \(AC = 20\) cm and angle \(BAC = 60^\circ\). Work out the length of \(AB\).

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   Model answer

   \(AB\) is adjacent to the \(60^\circ\) angle, so \(\cos 60^\circ = \dfrac{AB}{20}\). Since \(\cos 60^\circ = \dfrac{1}{2}\), \(AB = 10\) cm.

   Mark scheme

   • \(\cos 60^\circ = \dfrac{AB}{20}\) — M1
   • \(\dfrac{1}{2} = \dfrac{AB}{20}\) — M1
   • 10 cm — A1
6. 6 Work out [4 marks]

   A right-angled triangle has an angle of \(45^\circ\) and a hypotenuse of 12 cm. (a) Show that the length of the side opposite the \(45^\circ\) angle is \(6\sqrt{2}\) cm. (2 marks) (b) Work out the area of the triangle. (2 marks)

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   Model answer

   (a) \(\sin 45^\circ = \dfrac{x}{12}\), so \(x = 12 \times \dfrac{\sqrt{2}}{2} = 6\sqrt{2}\). (b) The triangle is isosceles, so both shorter sides are \(6\sqrt{2}\). The area is \(\dfrac{1}{2} \times 6\sqrt{2} \times 6\sqrt{2} = \dfrac{1}{2} \times 72 = 36\) cm\(^2\).

   Mark scheme

   • (a) \(\sin 45^\circ = \dfrac{x}{12}\) and \(12 \times \dfrac{\sqrt{2}}{2}\) — M1
   • (a) \(6\sqrt{2}\) shown — C1
   • (b) \(\dfrac{1}{2} \times 6\sqrt{2} \times 6\sqrt{2}\) — M1
   • (b) 36 cm\(^2\) — A1

Quick check

1. 1

   What is the formula for \(\sin\theta\) in a right-angled triangle?

   1. A\(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\)
   2. B\(\dfrac{\text{Opposite}}{\text{Hypotenuse}}\)
   3. C\(\dfrac{\text{Opposite}}{\text{Adjacent}}\)
   4. D\(\dfrac{\text{Hypotenuse}}{\text{Opposite}}\)
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   B: \(\dfrac{\text{Opposite}}{\text{Hypotenuse}}\)

   SOH: sine is opposite over hypotenuse.

2. 2

   Which ratio links the opposite side and the adjacent side?

   1. ATangent
   2. BSine
   3. CCosine
   4. DPythagoras
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   A: Tangent

   TOA: tangent is opposite over adjacent.

3. 3

   What is the exact value of \(\sin 30^\circ\)?

   1. A\(\dfrac{\sqrt{3}}{2}\)
   2. B1
   3. C\(\dfrac{\sqrt{2}}{2}\)
   4. D\(\dfrac{1}{2}\)
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   D: \(\dfrac{1}{2}\)

   This is one of the exact values you must learn: \(\sin 30^\circ = \dfrac{1}{2}\).

4. 4

   What is the exact value of \(\tan 45^\circ\)?

   1. A0
   2. B\(\dfrac{1}{2}\)
   3. C1
   4. D\(\sqrt{3}\)
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   C: 1

   At \(45^\circ\) the opposite and adjacent sides are equal, so \(\tan 45^\circ = 1\).

5. 5

   A right-angled triangle has a hypotenuse of 8 cm and an angle of \(30^\circ\). What is the length of the side opposite the \(30^\circ\) angle?

   1. A16 cm
   2. B4 cm
   3. C\(4\sqrt{3}\) cm
   4. D2 cm
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   B: 4 cm

   \(\sin 30^\circ = \dfrac{x}{8}\), so \(x = \dfrac{1}{2} \times 8 = 4\).

6. 6

   A right-angled triangle has opposite side 5 cm and adjacent side 5 cm. What is the angle?

   1. A\(45^\circ\)
   2. B\(30^\circ\)
   3. C\(60^\circ\)
   4. D\(90^\circ\)
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   A: \(45^\circ\)

   \(\tan\theta = \dfrac{5}{5} = 1\), so \(\theta = 45^\circ\).

7. 7

   In a right-angled triangle, \(\sin\theta = \dfrac{5}{13}\) and the hypotenuse is 39 cm. What is the opposite side?

   1. A3 cm
   2. B195 cm
   3. C5 cm
   4. D15 cm
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   D: 15 cm

   \(x = \dfrac{5}{13} \times 39 = 5 \times 3 = 15\).

8. 8

   A right-angled triangle has an angle of \(60^\circ\) and an adjacent side of 5 cm. What is the opposite side? (\(\tan 60^\circ = \sqrt{3}\))

   1. A\(\dfrac{5\sqrt{3}}{3}\) cm
   2. B\(\dfrac{5}{2}\) cm
   3. C\(5\sqrt{3}\) cm
   4. D\(5\sqrt{2}\) cm
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   C: \(5\sqrt{3}\) cm

   \(\tan 60^\circ = \dfrac{x}{5}\), so \(x = 5\sqrt{3}\).

9. 9

   A right-angled triangle has an angle of \(45^\circ\) and a hypotenuse of 10 cm. What is the opposite side?

   1. A\(10\sqrt{2}\) cm
   2. B\(5\sqrt{2}\) cm
   3. C\(5\sqrt{3}\) cm
   4. D5 cm
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   B: \(5\sqrt{2}\) cm

   \(x = 10 \sin 45^\circ = 10 \times \dfrac{\sqrt{2}}{2} = 5\sqrt{2}\).