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Practise 19 exam-style questions on Exact trigonometric values for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
x is an acute angle and \(\sin x = \dfrac{1}{2}\)
Write down the value of x.
\(\sin 30^\circ = \dfrac{1}{2}\), so \(x = 30^\circ\)
You should know that \(\sin 30^\circ = \frac{1}{2}\) (the side opposite 30° in a half-equilateral triangle is half the hypotenuse). A common mistake is to answer 60°, because \(\cos 60^\circ = \frac{1}{2}\).
B1 for 30°
cao
Kim says, "The exact value of tan 45° is \(\dfrac{\sqrt{2}}{2}\)."
Kim is wrong.
Write down the exact value of tan 45° and explain the mistake Kim has made.
\(\tan 45^\circ = 1\)
Kim has given the value of \(\sin 45^\circ\) (which is also \(\cos 45^\circ\)). For tan, the opposite and adjacent sides of the isosceles right-angled triangle are equal, so \(\tan 45^\circ = 1\).
Using the right-angled isosceles triangle with shorter sides 1 and hypotenuse \(\sqrt{2}\): \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\), but \(\tan 45^\circ = \frac{1}{1} = 1\). Kim has mixed up the ratios.
B1 for tan 45° = 1
C1 for a correct explanation, e.g. \(\dfrac{\sqrt{2}}{2}\) is the value of sin 45° (or cos 45°), not tan 45°
oe, e.g. "in a 45° right-angled triangle the opposite and adjacent sides are equal, so tan 45° = 1"
The diagram shows a right-angled triangle PQR.
PQ = 6 cm and angle QPR = 45°.
Work out the exact length of PR.
Give your answer in the form \(a\sqrt{2}\), where a is an integer.
\(\cos 45^\circ = \dfrac{6}{PR}\)
\(PR = \dfrac{6}{\cos 45^\circ} = \dfrac{6}{\frac{\sqrt{2}}{2}} = \dfrac{12}{\sqrt{2}} = \dfrac{12\sqrt{2}}{2} = 6\sqrt{2}\) cm
Either use \(\cos 45^\circ = \frac{\sqrt{2}}{2}\) and rationalise the denominator, or notice that the triangle is isosceles (QR = 6 cm too) and use Pythagoras: \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\). A decimal (8.49 cm) is not in the form asked for.
M1 for \(\cos 45^\circ = \dfrac{6}{PR}\) with \(\cos 45^\circ = \dfrac{\sqrt{2}}{2}\) (or \(\dfrac{1}{\sqrt{2}}\))
oe; or QR = 6 (isosceles) and \(PR^2 = 6^2 + 6^2\); correct answer with no working scores full marks
M1 for \(PR = \dfrac{12}{\sqrt{2}}\) or \(PR = \sqrt{72}\)
oe
A1 for \(6\sqrt{2}\)
cao
ABCD is a rectangle.
The diagonal AC has length \(10\sqrt{3}\) cm and angle CAB = 30°.
The diagram shows the rectangle.
Work out the exact perimeter of the rectangle.
Give your answer in the form \(a + b\sqrt{3}\), where a and b are integers.
\(AB = 10\sqrt{3} \times \cos 30^\circ = 10\sqrt{3} \times \dfrac{\sqrt{3}}{2} = \dfrac{30}{2} = 15\) cm
\(BC = 10\sqrt{3} \times \sin 30^\circ = 10\sqrt{3} \times \dfrac{1}{2} = 5\sqrt{3}\) cm
Perimeter \(= 2 \times 15 + 2 \times 5\sqrt{3} = 30 + 10\sqrt{3}\) cm
The diagonal is the hypotenuse of the right-angled triangle ABC. Use cos 30° for the adjacent side AB and sin 30° for the opposite side BC; \(\sqrt{3} \times \sqrt{3} = 3\) makes AB a whole number. Remember the perimeter has two lengths and two widths.
M1 for \(AB = 10\sqrt{3} \times \cos 30^\circ\)
oe; correct answer with no working scores full marks
M1 for \(BC = 10\sqrt{3} \times \sin 30^\circ\)
oe
A1 for AB = 15 and \(BC = 5\sqrt{3}\)
A1 for \(30 + 10\sqrt{3}\)
cao
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