Worksheets · Foundation and Higher

Exact trigonometric values

8 exam-style questions, grades 4 to 7. Worked solutions and the marks are on the last page.

Question 1Non-calculator · 3 marks
Write down the exact value of each expression.
(a) $\sin 30^\circ$ (1)
(b) $\cos 60^\circ$ (1)
(c) $\tan 45^\circ$ (1)
Question 2Non-calculator · 3 marks
$\theta$ is an acute angle.
(a) $\cos\theta = \frac{\sqrt{3}}{2}$ . Write down the value of $\theta$ . (1)
(b) Work out the value of $\sin 90^\circ - \cos 0^\circ + \tan 45^\circ$ . (2)
Question 3Non-calculator · 4 marks
A right isosceles triangle has both shorter sides 3 cm.
(a) Find the hypotenuse exactly. (2)
(b) Hence work out the exact value of $\sin 45^\circ$ . (2)
Question 4Non-calculator · 4 marks
Find exact values without using a calculator.
(a) Work out $4\sin30^\circ+2\cos60^\circ$ . (2)
(b) Work out $\tan60^\circ\times\tan30^\circ$ . (2)
Question 5Non-calculator · 4 marks
In triangle PQR, angle Q = $90^\circ$ , angle P = $60^\circ$ and PQ = 5 cm.
(a) Find the exact length of PR. (2)
(b) Find the exact length of QR. (2)
Question 6Non-calculator · 5 marks
Work out each exact value without a calculator.
(a) Work out $\sin 60^\circ \times \cos 30^\circ$ . (2)
(b) Work out $\tan 45^\circ - 2\sin 30^\circ$ . (2)
(c) Write down the acute angle θ for which $\tan\theta = \sqrt{3}$ . (1)
Question 7Non-calculator · 1 mark
(a) Work out the exact value of tan $60^\circ$ $\times$ sin $30^\circ.$ Circle one answer. (1)
1. $\sqrt{3}$ /2
2. $\sqrt{3}$
3. 1/( $2\sqrt{3}$ )
4. 3/2
Question 8Non-calculator · 6 marks
In triangle $ABC$ , $D$ is the point on $BC$ such that $AD$ is perpendicular to $BC$ . $AD = 6$ cm, angle $ABD = 30^\circ$ and angle $ACD = 45^\circ$ .
(a) Work out the exact length of $BC$ . Give your answer in the form $a + b\sqrt{3}$ . (4)
(b) Work out the exact length of $AB$ . (2)

Worked solutions and marks

Question 1

(a) $\frac{1}{2}$

From half an equilateral triangle: opposite 1, hypotenuse 2.

B1 $\frac{1}{2}$ or 0.5.

(b) $\frac{1}{2}$

In the same triangle, the side next to $60^\circ$ is 1 and the hypotenuse 2.

B1 $\frac{1}{2}$ or 0.5.

(c) 1

A right-angled isosceles triangle has opposite $=$ adjacent, so $\tan 45^\circ = 1$ .

B1 Correct answer: 1.

Question 2

(a) $30^\circ$

$\cos 30^\circ = \frac{\sqrt{3}}{2}$ .

B1 Correct answer: $30^\circ$ .

(b) 1

$1 - 1 + 1 = 1$

M1 At least two of $\sin 90^\circ = 1$ , $\cos 0^\circ = 1$ , $\tan 45^\circ = 1$ .
A1 Correct answer: 1.

Question 3

(a) $3\sqrt{2}$

Apply Pythagoras to the two equal shorter sides.
$\sqrt{3^{2}+3^{2}}$
Therefore $3\sqrt{2}$ .

M1 Apply Pythagoras to the two equal shorter sides.
A1 Correct answer: $3\sqrt{2}$

(b) $\sqrt{2}/2$

Divide the opposite side by the hypotenuse.
$3/(3\sqrt{2})$
Therefore $\sqrt{2}/2$ .

M1 Divide the opposite side by the hypotenuse.
A1 Correct answer: $\sqrt{2}/2$

Question 4

(a) $3$

Use one half for both sine 30 and cosine 60.
$4\times 1/2+2\times 1/2$
Therefore $3$ .

M1 Use one half for both sine 30 and cosine 60.
A1 Correct answer: $3$

(b) $1$

The exact values are root 3 and 1 over root 3.
$\sqrt{3}/\sqrt{3}$
Therefore $1$ .

M1 The exact values are root 3 and 1 over root 3.
A1 Correct answer: $1$

Question 5

(a) $10$ cm

PQ is adjacent to angle P and PR is the hypotenuse.
$5/\cos(60)$
Therefore $10$ cm.

M1 PQ is adjacent to angle P and PR is the hypotenuse.
A1 Correct answer: $10$ cm

(b) $5\sqrt{3}$

QR is opposite angle P.
$5\times \tan(60)$
Therefore $5\sqrt{3}$ .

M1 QR is opposite angle P.
A1 Correct answer: $5\sqrt{3}$

Question 6

(a) $\frac{3}{4}$

Both values are √3/2.
$\sqrt{3}/2\sqrt{3}/2$
Therefore $\frac{3}{4}$ .

M1 Both values are √3/2.
A1 Correct answer: $\frac{3}{4}$

(b) $0$

Use tan 45° = 1 and sin 30° = 1/2.
$1-2\times 1/2$
Therefore $0$ .

M1 Use tan 45° = 1 and sin 30° = 1/2.
A1 Correct answer: $0$

(c) $60$ °

Therefore $60$ °.

B1 Correct answer: $60$ °

Question 7

(a) $\sqrt{3}$ /2

Use exact values tan $60^\circ$ = $\sqrt{3}$ and sin $30^\circ$ = 1/2.
Multiply: $\sqrt{3}$ $\times$ 1/2 = $\sqrt{3}$ /2.

B1 Correct answer: $\sqrt{3}$ /2

Question 8

(a) $6 + 6\sqrt{3}$ cm

In triangle $ABD$ : $\tan 30^\circ = \frac{6}{BD}$ .
$BD = \frac{6}{\tan 30^\circ} = \frac{6}{1/\sqrt{3}} = 6\sqrt{3}$
In triangle $ADC$ : $\tan 45^\circ = \frac{6}{DC}$ .
$DC = \frac{6}{1} = 6$
$BC = 6 + 6\sqrt{3}$

P1 $BD = \frac{6}{\tan 30^\circ}$ with an exact value for $\tan 30^\circ$ .
P1 $DC = 6$ from $\tan 45^\circ = 1$ .
P1 $BD = 6\sqrt{3}$ simplified.
A1 $6 + 6\sqrt{3}$ .

(b) 12 cm

$\sin 30^\circ = \frac{6}{AB}$ , so $AB = \frac{6}{1/2} = 12$ .

M1 $\frac{6}{\sin 30^\circ}$ .
A1 Correct answer: 12 cm.

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Exact trigonometric values

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