These values come from two triangles: half of an equilateral triangle of side 2 gives \(30^\circ\) and \(60^\circ\), and half of a square of side 1 gives \(45^\circ\). The values at \(0^\circ\) and \(90^\circ\) come from the graphs.

For \(45^\circ\): \(\sin 45^\circ = \frac{\sqrt{2}}{2}\) and \(\cos 45^\circ = \frac{\sqrt{2}}{2}\).

So \(\sin 45^\circ = \frac{\sqrt{2}}{2}\)

You may also see this written as \(\frac{1}{\sqrt{2}}\). It is the same number — multiplying top and bottom by \(\sqrt{2}\) turns one into the other — and either form is correct.

The whole row for \(30^\circ\) is

\(\sin 30^\circ = \frac{1}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\tan 30^\circ = \frac{\sqrt{3}}{3}\)

so the two missing entries are \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\tan 30^\circ = \frac{\sqrt{3}}{3}\).

The tangent is a useful check on the other two, because \(\tan x = \frac{\sin x}{\cos x}\): dividing the sine by the cosine must give the tangent in the same row.

The tangent is often written \(\frac{1}{\sqrt{3}}\) instead. That is the same number as \(\frac{\sqrt{3}}{3}\), and either form is correct.

Write down the two exact values first: \(\tan 45^\circ = 1\) and \(\cos 0^\circ = 1\).

Now substitute them in.

\(\tan 45^\circ + \cos 0^\circ = 1 + 1 = 2\)

This is one of the exact values that has to be known: \(\cos 90^\circ = 0\).

In the range \(0 \leq x \leq 90\) no other angle gives that value, so there is only one solution.

\(x = 90\)

This is one of the values that has to be known rather than looked up.

\(\sin 30^\circ = \frac{1}{2}\)

\(AC\) is the hypotenuse, because it is the side opposite the right angle at \(B\). \(BC\) is opposite the angle at \(A\), so this is the sine ratio.

\(\sin 30^\circ = \frac{BC}{ 12 }\)

\(BC = 12 \times \frac{1}{2}\)

\(BC = 6\) cm

"Exact" means the surd stays as it is — turning it into a decimal would lose the exactness the question is asking for.

Write down the exact values: \(\tan 60^\circ = \sqrt{3}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\).

Combine the two exact values, and the surds cancel:

\(\sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{3}{2}\)

The coefficients give a factor of \(4\), so this part of the expression is \(4 \times \frac{3}{2} = 6\).

The other part is \(2 \times \frac{1}{2} = 1\).

Add them: \(6 + 1 = 7\).

The value is \(7\), which is an integer.

Write down the exact values: \(\cos 60^\circ = \frac{1}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\sin 45^\circ = \frac{\sqrt{2}}{2}\).

Substitute them in: \(\frac{\cos 60^\circ \sin 60^\circ}{\sin 45^\circ} = \frac{\frac{1}{2} \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}\)

Multiply the two values on the top: \(\frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}\)

Divide by the value on the bottom: \(\frac{\sqrt{3}}{4} \div \frac{\sqrt{2}}{2} = \frac{\sqrt{3}}{4} \times \frac{2}{\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}}\)

There is still a surd in the denominator, so multiply the top and the bottom by that surd: \(\frac{\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3} \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{6}}{4}\)

This is in the form \(\frac{ \sqrt{ a } }{ b }\) with \(a = 6\) and \(b = 4\).

Use the exact values \(\sin 60^\circ = \frac{\sqrt{3}}{2}, \tan 60^\circ = \sqrt{3}\) and \(\cos 60^\circ = \frac{1}{2}\).

Numerator: \(10 \times \frac{\sqrt{3}}{2} - \sqrt{3} = \frac{ 10\sqrt{3} }{2} - \frac{ 2\sqrt{3} }{2} = \frac{ 8\sqrt{3} }{2}\)

Denominator: \(8 \times \frac{1}{2} = \frac{ 8 }{2}\)

Divide, the halves cancel: \(\dfrac{ 8\sqrt{3} }{ 8 } = \sqrt{3}\)

This is a value from the table of exact values: \(\tan 60^\circ = \sqrt{3}\).

So the expression can be written as \(\tan\, 60^\circ\). So \(x + 17 = 60\), giving \(x = 60 - 17 = 43\).

Draw a perpendicular from \(A\) and one from \(B\) down to \(DC\). Each makes a right-angled triangle with hypotenuse \(s = 4\sqrt{3}\) cm and an angle of \(60^\circ\). The two triangles are the same because the trapezium is isosceles.

Height, opposite the angle: \(h = 4\sqrt{3} \times \sin 60^\circ = 4\sqrt{3} \times \frac{\sqrt{3}}{2} = 6\) cm

Overhang, next to the angle: \(4\sqrt{3} \times \cos 60^\circ = 4\sqrt{3} \times \frac{1}{2} = 2\sqrt{3}\) cm

\(DC\) is the given side. There is one overhang at each end, so \(2 \times 2\sqrt{3}\) is taken from \(DC\) to reach \(AB\).

Sum of the parallel sides: \(AB + DC = 2 \times 9 - 2 \times 2\sqrt{3}\)

Area \(= \frac{1}{2}(AB + DC) \times h = (9 - 2\sqrt{3}) \times 6\)

Multiply out.

Area \(= 54 - 12\sqrt{3}\) cm²

Recognise each angle from its exact value.

\(\sin 30^\circ = \frac{1}{2}\), so \(x = 30^\circ\).

\(\tan 60^\circ = \sqrt{3}\), so \(y = 60^\circ\).

Substitute into \(w\).

\(w = 4 \times 30^\circ - 60^\circ = 120^\circ - 60^\circ = 60^\circ\)

Now use the exact value at \(60^\circ\).

\(4 \sin w = 4 \sin 60^\circ = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\)