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Pythagoras and Trigonometry in 2D

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen and paper handy, to work out your answers.

Question 11 mark

Do not use a calculator.

Which of these is the value of \dfrac{\sin 30^\circ}{\cos 45^\circ} ?

Select the correct answer.

Hint

Write down \sin 30^\circ and \cos 45^\circ from the 1:\sqrt3:2 and 1:1:\sqrt2 triangles, then divide.

Worked solution
  1. \sin 30^\circ=\dfrac12
  2. \cos 45^\circ=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}
  3. \dfrac12\div\dfrac{1}{\sqrt2}=\dfrac12\times\sqrt2=\dfrac{\sqrt2}{2}
  4. Answer: \dfrac{\sqrt2}{2}

Question 22 marks

Do not use a calculator.

An equilateral triangle has perpendicular height 9 cm.

Work out the exact length of one side of the triangle.

Give your answer in the form a\sqrt3, where a is an integer.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

The height splits the triangle into two right-angled triangles with a 60^\circ angle; use \sin 60^\circ=\dfrac{\sqrt3}{2}.

Worked solution
  1. The height cuts the triangle into two right-angled triangles with angles 30^\circ, 60^\circ, 90^\circ
  2. The side s is the hypotenuse and the height is opposite the 60^\circ angle
  3. \sin 60^\circ=\dfrac{9}{s}, so \dfrac{\sqrt3}{2}=\dfrac{9}{s}
  4. s=\dfrac{18}{\sqrt3}
  5. Rationalise: \dfrac{18}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{18\sqrt3}{3}=6\sqrt3
  6. Answer: 6\sqrt3 cm

Question 3Challenge6 marks

Do not use a calculator.

O is the origin and C is the point (12,\,0)

A is a point above the x-axis such that angle AOC=45^\circ and angle ACO=30^\circ

(a)

Work out the exact coordinates of A.

Give each coordinate in the form a\sqrt3+b, where a and b are integers.

4 marks

Write your answer as (x, y)

(b)

Work out the exact area of triangle OAC.

Give your answer in the form p\sqrt3+q, where p and q are integers.

2 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Drop a perpendicular from A to the x-axis and call its length h; write both parts of OC in terms of h.

Worked solution

Part (a)

  1. Let D be the point on the x-axis directly below A, and let AD=h
  2. Triangle AOD has a 45^\circ angle, so it is isosceles: OD=h
  3. Triangle ACD: \tan 30^\circ=\dfrac{h}{DC}, so DC=\dfrac{h}{\tan 30^\circ}=h\sqrt3
  4. OD+DC=12: h+h\sqrt3=12
  5. h=\dfrac{12}{1+\sqrt3}
  6. Rationalise: \dfrac{12}{1+\sqrt3}\times\dfrac{\sqrt3-1}{\sqrt3-1}=\dfrac{12(\sqrt3-1)}{3-1}=6\sqrt3-6
  7. A is h across and h up from O
  8. Answer: (6\sqrt3-6,\ 6\sqrt3-6)

Part (b)

  1. Base OC=12 and perpendicular height = the y-coordinate of A=6\sqrt3-6
  2. Area =\dfrac12\times12\times(6\sqrt3-6)
  3. =6(6\sqrt3-6)
  4. Answer: 36\sqrt3-36

More on this topic: Pythagoras and Trigonometry in 2D worksheet with full solutions

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