\(x\) is opposite the angle of \(40^\circ\), and the hypotenuse is \(8\) cm, so use sine.

\(\sin 40^\circ = \frac{x}{ 8 }\)

\(x = 8 \times \sin 40^\circ = 5.1\)

\(x = 5.1\) cm

The two sides given are opposite and adjacent to \(x\), so use tangent.

\(\tan x = \frac{ 4 }{ 5 }\)

\(x = \tan^{-1}\left( \frac{ 4 }{ 5 } \right) = 38.7^\circ\)

The angles of a triangle add to \(180^\circ\), and one of them is the right angle.

\(180 - 90 - 38.7 = 51.3\)

The third angle is \(51.3^\circ\).

Sketch the rectangle with its diagonal. The diagonal splits it into a right-angled triangle, and the diagonal of \(23.5\) cm is the hypotenuse.

\(x\) is the side adjacent to the angle of \(37^\circ\), and the hypotenuse is known, so use cosine.

\(\cos 37^\circ = \frac{ x }{ 23.5 }\)

\(x = 23.5 \times \cos 37^\circ \approx 18.8\)

So \(x\) is 18.8 cm, correct to 1 decimal place.

\(x\) is the hypotenuse and \(15\) cm is adjacent to the angle of \(54^\circ\), so use cosine.

\(\cos 54^\circ = \frac{ 15 }{ x }\)

\(x = \frac{ 15 }{\cos 54^\circ} = 25.5\)

\(x = 25.5\) cm

\(x\) is adjacent to the angle of \(60^\circ\), and the hypotenuse is \(22\) cm.

\(\cos 60^\circ = \frac{1}{2}\), so

\(x = 22 \times \frac{1}{2} = 11\)

\(x = 11\) cm

The height is opposite the angle of \(34^\circ\) and the ramp is the hypotenuse, so use sine.

\(h = 14 \times \sin 34^\circ = 7.8\)

The height is \(7.8\) m.

The horizontal distance is adjacent to the angle of \(34^\circ\), so use cosine.

\(d = 14 \times \cos 34^\circ = 11.6\)

The horizontal distance is \(11.6\) m.

Label the sides from the angle of \(32^\circ\). The side marked \(x\) is the opposite and the side of 20 cm is the adjacent.

So the correct equation is \(\mathrm{ tan }\,32^\circ = \frac{ x }{ 20 }\), which gives \(x = 20 \ × \ \mathrm{ tan }\,32^\circ\).

Compare each line of Sam's working with this.

The fraction should be x over 20, and Sam has it the other way up.

x is a length, so the angle goes straight into the ratio. An inverse function is only for finding an angle.

So the errors are statements \(2\) and \(6\).

The dashed line is opposite the \(43^\circ\) angle and the other side of the triangle on the outline is adjacent to it, so use tan.

Let \(x\) cm be the missing length. It is the height of the rectangle, which is the dashed line.

\(\tan 43^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{ x }{ 8 }\)

Rearranging for \(x\):

\(x = 8 × \tan 43^\circ \approx 7.46\text{ cm}\)

The dashed line is a side of the rectangle as well, so the two parts share it. From here use the unrounded value of \(x\) from your calculator, not \(7.46\).

\(\text{triangle} = \frac{1}{2} \times 8 \times x \approx 29.84\)

\(\text{rectangle} = x \times 4 \approx 29.84\)

Add the two areas, using the unrounded values.

\(\text{whole shape} = \text{triangle} + \text{rectangle} \approx 59.7\)

The area is \(59.7\) cm².

The side \(BD\) is in both triangles, and the given side of triangle \(CBD\) is not enough on its own, so find \(BD\) first.

In triangle \(ABD\), \(AD\) is adjacent to the marked angle and \(BD\) is opposite it, so use tangent.

\(BD = 9 \times \tan 51^\circ \approx 11.114\) cm

Now use triangle \(CBD\) with \(BD\) and the given side \(DC\).

In triangle \(CBD\), \(BD\) is opposite angle \(BCD\) and \(DC\) is adjacent to it, so use tangent.

\(\tan \angle BCD = \frac{ 11.114 }{ 16 }\)

\(\angle BCD = \tan^{-1}\left( \frac{ 11.114 }{ 16 } \right) \approx 34.8^\circ\)

(Use the unrounded value of \(BD\) on the calculator.)

\(AD\) is in triangle \(ADC\), but the given length is in triangle \(ABC\). They share the side \(AC\), so find \(AC\) first.

In triangle \(ABC\), \(AC\) is opposite the angle of \(33^\circ\) and \(CB\) is given, so use tangent.

\(\tan 33^\circ = \frac{ AC }{ CB }\)

\(AC = 13 \times \tan 33^\circ = 8.442\) cm

In triangle \(ADC\), \(AC\) is opposite the angle of \(50^\circ\) and \(AD\) is wanted, so use sine.

\(\sin 50^\circ = \frac{ 8.442 }{ AD }\)

\(AD = \frac{ 8.442 }{ \sin 50^\circ } = 11.0\) cm