Question: In a right-angled triangle, one angle is $52°$. The side next to this angle is $8$ cm and the hypotenuse is $x$ cm. Find $x$.

✏️ Sketch this first: right-angled triangle, 52° at the base, 8 cm adjacent, hypotenuse labelled x

Step 1. Label the sides as seen from the $52°$ angle.

$8$ cm touches the angle → adjacent.

$x$ faces the right angle → hypotenuse.

Adjacent and hypotenuse → use cosine (the CAH in TOA CAH SOH).

Step 2. Form the ratio:

$$\cos 52° = \dfrac{8}{x}$$

Step 3. The unknown is at the bottom of the fraction.

Multiply both sides by $x$, then divide both sides by $\cos 52°$:

$$x = \dfrac{8}{\cos 52°}$$

Step 4. Calculate, then round:

$$x = 12.9941\ldots \approx 13.0 \text{ cm} \quad (3 \text{ s.f.})$$

Check. The hypotenuse must be the longest side of the triangle, and $13.0 > 8$. ✓

⚠ Watch out: The hypotenuse always faces the right angle, so it sits on the bottom of CAH: $\cos 52° = \frac{8}{x}$, never $\frac{x}{8}$ — and when the unknown is underneath, you divide by $\cos 52°$, not multiply.

Question: In a right-angled triangle, the side opposite the angle $\theta$ is $4.1$ cm and the side adjacent to $\theta$ is $3.2$ cm. Find $\theta$.

✏️ Sketch this first: right-angled triangle with θ at one corner, 4.1 cm opposite it and 3.2 cm along the base

Step 1. Label the two known sides from $\theta$.

$4.1$ cm faces the angle → opposite.

$3.2$ cm touches the angle → adjacent.

Opposite and adjacent → use tangent (TOA).

Step 2. Form the ratio:

$$\tan\theta = \dfrac{4.1}{3.2}$$

Step 3. To get the angle out, apply the inverse:

$$\theta = \tan^{-1}\left(\dfrac{4.1}{3.2}\right)$$

Step 4. Calculate, then round:

$$\theta = 52.0283\ldots° \approx 52.0° \quad (1 \text{ d.p.})$$

Check. The opposite side ($4.1$) is longer than the adjacent ($3.2$), so $\theta$ should be more than $45°$. ✓

⚠ Watch out: Finding an angle needs the inverse key: $\theta = \tan^{-1}\frac{4.1}{3.2}$, not $\tan\frac{4.1}{3.2}$ — and remember lengths round to 3 s.f. but angles round to 1 d.p.

Question: In triangle $ABC$, $AB = 7$ cm, $BC = 24$ cm and $AC = 25$ cm. (a) Show that the triangle is right-angled. (b) $BA$ is produced to $D$. Write down $\cos\angle CAD$ and $\sin\angle CAD$.

Step 1. Test the sides against Pythagoras — square the two shorter sides and compare with the longest:

$$7^2 + 24^2 = 49 + 576 = 625, \qquad 25^2 = 625$$

Step 2. Since $AB^2 + BC^2 = AC^2$, the converse of Pythagoras' theorem says the triangle is right-angled — at $B$, the vertex facing the longest side. (Shown.)

Step 3. (b) $\angle CAD$ sits on the straight line $DAB$, so it is the obtuse partner of $\angle CAB$:

$$\angle CAD = 180° - \angle CAB$$

From the right-angled triangle: $\cos\angle CAB = \dfrac{7}{25}$, $\;\sin\angle CAB = \dfrac{24}{25}$.

Step 4. Apply the obtuse rules — sin keeps its sign, cos flips:

$$\cos\angle CAD = -\dfrac{7}{25}, \qquad \sin\angle CAD = \dfrac{24}{25}$$

⚠ Watch out: The right angle is at $B$, opposite the hypotenuse $AC$ — not at $A$. State the converse conclusion explicitly ("since $7^2+24^2 = 25^2$, $\angle ABC = 90°$"); computing $625 = 625$ without the sentence doesn't earn the "show" mark.

Question: In triangle $ABC$, $\angle A = 85°$, $\angle B = 38°$ and $AB = 10$ cm. Find $AC$.

Step 1. The sine rule pairs each angle with the side facing it. $AB$ faces $\angle C$ — which we don't have yet. Angles in a triangle:

$$\angle C = 180° - 85° - 38° = 57°$$

Step 2. Now we have 2 angles and 2 sides (counting the unknown): sine rule.

$AC$ faces $\angle B = 38°$; $AB = 10$ faces $\angle C = 57°$:

$$\dfrac{AC}{\sin 38°} = \dfrac{10}{\sin 57°}$$

Step 3. Make $AC$ the subject:

$$AC = \dfrac{10\sin 38°}{\sin 57°}$$

Step 4. Calculate, then round:

$$AC = 7.3409\ldots \approx 7.34 \text{ cm} \quad (3 \text{ s.f.})$$

Check. $AC$ faces the smallest angle used ($38° < 57°$), so it should be shorter than $10$ cm. ✓

The same setup finds any other side: pair it with its facing angle and solve.

⚠ Watch out: The sine rule pairs an angle with the side facing it — $\frac{AC}{\sin 38°}$, never $\frac{AC}{\sin 85°}$. If the facing angle isn't given, find it first with the angle sum.

Question: In triangle $ABC$, $\angle A = 18°$, $BC = 15$ cm and $AC = 20$ cm. The diagram shows that $\angle B$ is obtuse. Find $\angle B$.

✏️ Sketch this first: triangle with the 18° angle at A and an obtuse angle at B

Step 1. Two angles and two sides involved ($\angle A$ with $BC$, $\angle B$ with $AC$) → sine rule:

$$\dfrac{\sin\angle B}{20} = \dfrac{\sin 18°}{15}$$

Step 2. Make $\sin\angle B$ the subject:

$$\sin\angle B = \dfrac{20\sin 18°}{15} = 0.41202\ldots$$

Step 3. Apply $\sin^{-1}$ — and remember sine gives two possible angles between $0°$ and $180°$:

$$\angle B = 24.3319\ldots° \quad \text{or} \quad 180° - 24.3319\ldots° = 155.668\ldots°$$

Step 4. Decide using the diagram. $\angle B$ is obtuse, so reject the acute value:

$$\angle B \approx 155.7° \quad (1 \text{ d.p.})$$

Check. $155.7° + 18° = 173.7° < 180°$, so the triangle is still valid. ✓

⚠ Watch out: $\sin^{-1}$ on a calculator only ever gives the acute answer — the obtuse partner $180° - \theta$ must be written down and tested yourself. State the rejected value, then reject it.

Question: In triangle $ABC$, $AB = 5$ cm, $AC = 7$ cm and $\angle A = 80°$. Find $BC$.

Step 1. Count what you have: 1 angle and 3 sides involved (the unknown $BC$ counts as a side) → cosine rule.

The side you want, $BC$, must face the angle you use, $\angle A$.

Step 2. Write the cosine rule with $BC$ as the subject side:

$$BC^2 = 5^2 + 7^2 - 2(5)(7)\cos 80°$$

Step 3. Evaluate the right side in one go, then square-root:

$$BC^2 = 25 + 49 - 70\cos 80° = 61.8445\ldots$$ $$BC = \sqrt{61.8445\ldots} = 7.8641\ldots$$

Step 4. Round:

$$BC \approx 7.86 \text{ cm} \quad (3 \text{ s.f.})$$

Check. $\angle A = 80°$ is close to $90°$, so $BC$ should be close to (slightly under) the "Pythagoras" value $\sqrt{5^2+7^2} \approx 8.6$. ✓

⚠ Watch out: In $a^2 = b^2 + c^2 - 2bc\cos A$, the angle $A$ must be between the two known sides, and the answer side faces it. Keep $-2(5)(7)\cos 80°$ as one term — don't compute $25 + 49 - 70$ first and then multiply by $\cos 80°$.

Question: In triangle $PQR$, $PQ = 8$ cm, $QR = 11$ cm and $PR = 6$ cm. Find $\angle PQR$.

Step 1. All three sides known, angle wanted → cosine rule.

The angle $\angle PQR$ is at $Q$, so the side facing it is $PR = 6$.

Step 2. Substitute into $a^2 = b^2 + c^2 - 2bc\cos A$ with the facing side on the left:

$$6^2 = 8^2 + 11^2 - 2(8)(11)\cos\angle PQR$$

Step 3. Rearrange for the cosine in one step:

$$\cos\angle PQR = \dfrac{6^2 - 8^2 - 11^2}{-2(8)(11)}$$

Step 4. Evaluate, then apply the inverse:

$$\cos\angle PQR = \dfrac{36 - 64 - 121}{-176} = \dfrac{-149}{-176} = 0.84659\ldots$$ $$\angle PQR = \cos^{-1}(0.84659\ldots) = 32.157\ldots° \approx 32.2° \quad (1 \text{ d.p.})$$

Check. $\angle PQR$ faces the shortest side (6 cm), so it should be the smallest angle of the triangle — an answer under $60°$ makes sense. ✓

⚠ Watch out: The rearrangement is $\cos A = \frac{a^2 - b^2 - c^2}{-2bc}$ (equivalently $\frac{b^2 + c^2 - a^2}{2bc}$) — the facing side $a$ is the one that appears alone. Unlike $\sin^{-1}$, $\cos^{-1}$ needs no second case: it already tells you if the angle is obtuse (the cosine comes out negative).

Question: In the diagram, $\angle BAC = 95°$, $\angle ACB = 52°$, $AC = 5$ cm, $CD = 6$ cm and $BCD$ is a straight line. Find $AB$, $AD$ and the area of $\triangle ABD$.

✏️ Sketch this first: triangle ABC and triangle ACD sharing the side AC, with B, C, D on one line

Step 1. $AB$ lives in $\triangle ABC$: 2 angles ($95°$, $52°$) and 2 sides ($5$ and the unknown) → sine rule.

$$\angle ABC = 180° - 95° - 52° = 33°$$ $$\dfrac{AB}{\sin 52°} = \dfrac{5}{\sin 33°}$$ $$AB = \dfrac{5\sin 52°}{\sin 33°}$$ $$AB = 7.2342\ldots \approx 7.23 \text{ cm}$$

Step 2. $AD$ lives in $\triangle ACD$: 1 angle and 3 sides → cosine rule.

$\angle ACD = 180° - 52° = 128°$ (angles on a straight line).

$$AD^2 = 5^2 + 6^2 - 2(5)(6)\cos 128°$$ $$AD = 9.8964\ldots \approx 9.90 \text{ cm}$$

Step 3. Area of $\triangle ABD$ = area of $\triangle ABC$ + area of $\triangle ACD$:

$$\text{Area} = \dfrac{1}{2}(5)(7.2342\ldots)\sin 95° + \dfrac{1}{2}(5)(6)\sin 128°$$ $$\text{Area} = 29.8 \text{ cm}^2 \quad (3 \text{ s.f.})$$

⚠ Watch out: Count the information inside the triangle you're using: 2 angles + 2 sides → sine rule; 1 angle + 3 sides → cosine rule (the unknown counts as a side). And carry $7.2342\ldots$, not the rounded $7.23$, into the area — early rounding shifts the final answer.

Question: In triangle $ABC$, $AB = 12$ cm, $BC = 15$ cm and $\angle ABC = 108°$. Find the area of the triangle.

Step 1. Check the setup: two sides and the angle between them.

$AB$ and $BC$ both start at $B$, and $\angle ABC$ sits at $B$ — so $108°$ is the included angle. The formula applies directly:

$$\text{Area} = \dfrac{1}{2}ab\sin C$$

Step 2. Substitute the two sides and the included angle:

$$\text{Area} = \dfrac{1}{2}(12)(15)\sin 108°$$

Step 3. Evaluate:

$$\text{Area} = 90 \times \sin 108° = 85.595\ldots$$

Step 4. Round, with units:

$$\text{Area} \approx 85.6 \text{ cm}^2 \quad (3 \text{ s.f.})$$

Check. The obtuse angle doesn't break the formula — $\sin 108°$ is positive (about $0.95$), so the area is close to, but slightly less than, $\frac{1}{2}(12)(15) = 90$, the value a right angle would give. ✓

⚠ Watch out: The angle in $\frac{1}{2}ab\sin C$ must be the one between the two sides you used. Pairing $12$ and $15$ with an angle at $A$ or $C$ instead of $\angle ABC$ gives a wrong answer that still looks reasonable — always check the angle sits where the two sides meet.

Question: A boat is at $B$, $20$ km due north of the harbour $A$. It sails from $B$ on a bearing of $110°$ at $12$ km/h for $2$ hours, anchoring at $C$. Find $BC$, the distance $CA$, and the bearing of $A$ from $C$.

Step 1. Distance = speed × time:

$$BC = 12 \times 2 = 24 \text{ km}$$

Step 2. Find the angle inside the triangle at $B$.

$A$ is due south of $B$, so the bearing of $A$ from $B$ is $180°$. The boat leaves on $110°$:

$$\angle ABC = 180° - 110° = 70°$$

Step 3. Two sides and the included angle → cosine rule for $CA$:

$$CA^2 = 20^2 + 24^2 - 2(20)(24)\cos 70°$$ $$CA = 25.449\ldots \approx 25.4 \text{ km} \quad (3 \text{ s.f.})$$

Step 4. Bearing of $A$ from $C$: work from the bearing of $B$ from $C$, which is the back bearing $110° + 180° = 290°$.

Find $\angle BCA$ by sine rule:

$$\sin\angle BCA = \dfrac{20\sin 70°}{25.449\ldots}$$ $$\angle BCA = 47.60\ldots°$$

Turning from $CB$ towards $CA$ is anticlockwise, so subtract:

$$\text{Bearing of } A \text{ from } C = 290° - 47.60\ldots° \approx 242.4°$$

⚠ Watch out: A back bearing is $\pm 180°$, so the bearing of $B$ from $C$ is $110° + 180° = 290°$, not $360° - 110°$. Draw the north arrow at every point — the north arrows are parallel, which is what makes interior angles work.

Question: $ABCDEFGH$ is a cuboid with horizontal base $ABCD$, and $G$ is the vertex directly above $C$. Given $AB = 15$ cm, $BC = 8$ cm and the diagonal $AG = 20$ cm, find the height $GC$ and the angle $AG$ makes with the base ($\angle GAC$).

✏️ Sketch this first: cuboid with base ABCD, top EFGH, diagonal AG drawn

Step 1. Work in flat right-angled triangles, one at a time.

In the base, $\triangle ABC$ has a right angle at $B$:

$$AC = \sqrt{15^2 + 8^2} = \sqrt{289} = 17 \text{ cm}$$

Step 2. $GC$ is vertical, so it meets the base line $AC$ at $90°$. In $\triangle ACG$, $AG = 20$ is the hypotenuse:

$$GC = \sqrt{20^2 - 17^2} = \sqrt{111}$$ $$GC = 10.535\ldots \approx 10.5 \text{ cm} \quad (3 \text{ s.f.})$$

Step 3. The angle between $AG$ and the base is the angle between $AG$ and its shadow on the base, $AC$ — that is $\angle GAC$ in the same triangle:

$$\cos\angle GAC = \dfrac{AC}{AG} = \dfrac{17}{20}$$

Step 4. Apply the inverse:

$$\angle GAC = \cos^{-1}\left(\dfrac{17}{20}\right)$$ $$\angle GAC = 31.788\ldots° \approx 31.8° \quad (1 \text{ d.p.})$$

⚠ Watch out: The space diagonal is $AG = \sqrt{AB^2 + BC^2 + CG^2}$ — the face diagonal $\sqrt{AB^2 + BC^2}$ is only $AC$. To find the angle with the base, always project onto the base first: use $AC$ (the shadow), and note $AG$ is the hypotenuse, so $\cos\angle GAC = \frac{AC}{AG}$, not $\frac{AC}{CG}$.

Question: $ABCDEF$ is a right triangular prism. $\angle ABC = 90°$, $\angle ACB = 18°$, $AC = 30$ cm and $AD = 80$ cm ($AD$ is a horizontal edge; $CF = AD$). Find $AB$, $\angle CAF$, and $\angle AFB$.

✏️ Sketch this first: triangular prism with vertical triangle ABC at one end, F below C's far end

Step 1. $AB$ lives in the end triangle $ABC$:

$$\sin 18° = \dfrac{AB}{30}$$ $$AB = 30\sin 18°$$ $$AB = 9.2705\ldots \approx 9.27 \text{ cm}$$

Step 2. $\triangle ACF$ is right-angled with $CF = AD = 80$:

$$\tan\angle CAF = \dfrac{80}{30}$$ $$\angle CAF = 69.4° \quad (1 \text{ d.p.})$$

Step 3. For $\angle AFB$, first get $AF$ from right-angled $\triangle ADF$ ($DF = AC = 30$):

$$AF = \sqrt{80^2 + 30^2} = 85.440\ldots$$

Step 4. $\triangle ABF$ is right-angled at $B$:

$$\sin\angle AFB = \dfrac{AB}{AF} = \dfrac{9.2705\ldots}{85.440\ldots}$$ $$\angle AFB = 6.2° \quad (1 \text{ d.p.})$$

⚠ Watch out: Every 3-D answer here came from a flat right-angled triangle — the skill is choosing which one, and carrying unrounded values ($9.2705$, $85.440$) between them. Using the rounded $9.27/85.4$ shifts the final angle. Identify each triangle's right angle before writing any ratio.

Question: $P$, $S$ and $Q$ lie on a straight line in that order. $R$ is a point above the line with $RQ \perp PQ$, $RQ = 15$ cm and $SQ = 8$ cm. Find $SR$, $\sin\angle PSR$ and $\cos\angle PSR$, giving the ratios as fractions.

✏️ Sketch this first: right-angled triangle RSQ standing on the line, with the base extended through S to P

Step 1. Find $SR$ using Pythagoras in $\triangle SQR$:

$$SR = \sqrt{15^2 + 8^2} = \sqrt{289} = 17 \text{ cm}$$

Step 2. Spot that $\angle PSR$ is obtuse — it sits outside the triangle, on a straight line with the acute angle $\angle RSQ$:

$$\angle PSR = 180° - \angle RSQ$$

Step 3. Read off the acute ratios from $\triangle SQR$ first:

$$\sin\angle RSQ = \dfrac{15}{17}, \qquad \cos\angle RSQ = \dfrac{8}{17}$$

Step 4. Apply the obtuse-angle rules: $\sin(180° - \theta) = \sin\theta$ but $\cos(180° - \theta) = -\cos\theta$.

$$\sin\angle PSR = \dfrac{15}{17}, \qquad \cos\angle PSR = -\dfrac{8}{17}$$

⚠ Watch out: Only cos (and tan) change sign for an obtuse angle — $\sin(180° - \theta) = \sin\theta$, not $-\sin\theta$. Writing $\cos\angle PSR = \frac{8}{17}$ without the minus is the classic mark-loser.

Question: A vertical tree $CT$ stands with its base at $C$. The point $A$ is on level ground, $72$ m from $C$. The angle of elevation of the top of the tree $T$ from $A$ is $15°$. Find the height of the tree.

Step 1. Draw the right-angled triangle.

The ground $AC$ is horizontal, the tree $CT$ is vertical, so $\angle ACT = 90°$.

The angle of elevation is measured from the horizontal at $A$ up to the line of sight $AT$: $\angle TAC = 15°$.

Step 2. Label from the $15°$ angle: $CT$ is opposite, $AC = 72$ is adjacent → tangent.

$$\tan 15° = \dfrac{CT}{72}$$

Step 3. Make $CT$ the subject:

$$CT = 72\tan 15°$$

Step 4. Calculate and round:

$$CT = 19.292\ldots \approx 19.3 \text{ m} \quad (3 \text{ s.f.})$$

Check. A $15°$ elevation is shallow, so the height should be much less than the $72$ m distance. ✓

⚠ Watch out: Elevation and depression are both measured from the horizontal, never from the vertical — and the angle of depression from $T$ down to $A$ equals the angle of elevation from $A$ up to $T$ (alternate angles), not $90°$ minus it.