Right-angled Trigonometry Worked ExamplesIGCSE Maths 0580

Choose the ratio from which two sides are involved, not by habit. When candidates found a side in a right-angled triangle, the June 2025 examiner report flagged "the error of giving their answer to 2 or fewer significant figures", so write down $6.883\ldots$ and then round to $6.88$. Sense-check: the opposite side must be shorter than the hypotenuse, and $6.88 < 12$. ✓ Make sure your calculator is in degrees.

Keep the fraction inside the inverse function — computing $7 \div 9 = 0.78$ first and then inverting loses accuracy. Since the opposite is shorter than the adjacent, $\theta$ must be less than $45^{\circ}$, which it is. ✓

In (b) use the two given sides, not your calculated hypotenuse — that way an error in (a) cannot carry through. The largest angle is always opposite the longest side, which here is the $90^{\circ}$ angle opposite the hypotenuse.

In (b), using $6.5\sin 72.1^{\circ}$ also works but relies on your rounded answer from (a). Where an exact route exists, take it. The June 2025 examiner report stressed that "intermediate values within longer methods should not be rounded, with only the final answer rounded to the appropriate degree of accuracy."

The angle of depression is measured from the horizontal, not from the vertical cliff face. Using $68^{\circ}$ (the complement) is the standard error and gives $198 \to 32.3$. A quick sketch with the horizontal dashed line prevents it.

The two triangles share the height, which is what lets you eliminate the unknown distance. Note $\tan 45^{\circ} = 1$ exactly — recognising that simplifies the algebra. Keep at least four figures in the bracket: the March 2026 examiner report reminded candidates to "use more than 3 figures in their intermediate calculations". Check: $\frac{45.4}{\tan 28^{\circ}} \approx 85.4$ and $\frac{45.4}{1} = 45.4$, differing by $40$. ✓

Mark the right angle in each triangle before choosing a ratio. A common error in (c) is to treat $AC$ as the hypotenuse and write $AD = 10\cos 45^{\circ} = 5\sqrt{2}$, which would make $AD$ shorter than $AC$. That cannot happen, because the hypotenuse is always the longest side. The June 2025 examiner report advised that "it is useful to know the trigonometric ratios for the angles 0°, 30°, 45°, 60° and 90°, with the exception of tan90°." Here you need: $\sin 30^{\circ} = \cos 60^{\circ} = \tfrac{1}{2}$, $\cos 30^{\circ} = \sin 60^{\circ} = \tfrac{\sqrt{3}}{2}$ and $\sin 45^{\circ} = \cos 45^{\circ} = \tfrac{1}{\sqrt{2}}$. A decimal such as $8.66$ does not earn the mark when the question asks for the form $k\sqrt{3}$.

In a show that part, the marks are for the working and a more accurate value ($9.5966\ldots$) before rounding. The March 2025 examiner report noted that a bare answer "could not gain the marks on its own since this was easy to work backwards to reach", and the March 2026 report added that "they should show every step in their method". Both angles are measured from the same horizontal line through $W$, so the building splits into two right-angled triangles. A common error is to add the angles and work out $25\tan 53^{\circ} = 33.2$, which is not a length in either triangle. In (c), the angle of elevation from ground level is larger than $32^{\circ}$ because $P$ is lower than $W$.

The whole question depends on seeing the right-angled triangle $OMA$ and knowing which length you have. The $75\text{ cm}$ is measured across the flats, so $37.5\text{ cm}$ is $OM$, the perpendicular from the centre. A common error is to treat $37.5$ as $OA$, the distance to a vertex, and use sine instead. Keep the side unrounded in (c) and (d): using $31.1$ gives $4665\text{ cm}^2$ and $17.1\%$, and the final accuracy mark is lost. As the March 2025 examiner report put it, "intermediate values within longer methods should not be rounded".

The June 2025 examiner report found that "it was evident that many candidates did not have a method of deriving tan60 without a calculator." The half-equilateral triangle in (b) is that method, and it gives every exact ratio for $30^{\circ}$ and $60^{\circ}$. On a question asking for an exact side, the same report noted that some "candidates assumed that the triangle had side lengths in the ratio 3 : 4 : 5". A $30^{\circ}$ angle never gives a $3 : 4 : 5$ triangle. Do not swap the values at $90^{\circ}$: $\sin 90^{\circ} = 1$ but $\cos 90^{\circ} = 0$. In (c), $\frac{12}{\sqrt{3}}$ is correct but not in the form asked for. Check: $4\sqrt{3} \approx 6.93$, which is less than $12$, as it must be for the side opposite the smallest angle.

The March 2026 examiner report found that "Quite a few candidates were not able to correctly rationalise the denominator of the surd". Multiply the numerator and the denominator by $\sqrt{3}$ and then simplify. Part (b) asks for the perpendicular distance from $R$ to $PQ$. On a question like this, the same report noted that quite a lot of candidates drew a line from the right-angled vertex to the hypotenuse and "assumed that it bisected the angle". The perpendicular does not bisect the right angle here. You can check (b) with areas: $QR = \frac{9}{\tan 60^{\circ}} = 3\sqrt{3}$, and $\tfrac{1}{2} \times 9 \times 3\sqrt{3} = \tfrac{1}{2} \times 6\sqrt{3} \times RN$ gives $RN = 4.5$. ✓

The June 2025 examiner report found that "Most candidates did not recognise that the shortest distance was found by using a perpendicular". Once you draw the perpendicular, part (a) is a two-step right-angled triangle calculation. The same report said a common error was to draw a line that bisected the angle, and the other was to join the vertex to the midpoint of the opposite side. Neither of those lines meets the path at a right angle. The $104^{\circ}$ angle is not needed, because it is not in triangle $DNE$.

On a question with two joined right-angled triangles, the June 2025 examiner report found that many candidates wrongly took two sides in different triangles to be equal, "possibly interpreting both triangles as being congruent". The triangles share only $FH$, and $FH$ plays a different role in each: it is the hypotenuse of $FGH$ but the adjacent side of $FHJ$. Mark the right angle in each triangle before choosing a ratio. Check with $k = 1$: $FH = \sqrt{2} \approx 1.414$ and $y \approx 1.414 \times 1.732 = 2.449 \approx \sqrt{6}$. ✓