Mixed Triangle Problems (AA SL)
Real exam questions rarely announce which triangle technique they want. A cliff-and-boat scenario might need right-angle trigonometry for one leg and nothing more; a ladder against a leaning wall might need the angle sum of a triangle followed by the sine rule. This page is about spotting which tool fits which stage of a multi-step problem. It's part of the broader Triangles (Sine/Cosine Rules) topic.
14 questions on this sub-topic.
Choosing the right tool
These questions sit under IB syllabus reference SL3.2, which covers the sine rule, the cosine rule and triangle area - but a mixed problem often needs right-angle trigonometry too, before the triangle-specific rules even come into play.
Right-angle stage
\(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\), \(a^2+b^2=c^2\)
Prior-knowledge tools, not in the formula booklet. Reach for these the moment a right angle appears - heights, cliffs, 3D solids, angles of elevation/depression.
No right angle
\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\), \(c^2=a^2+b^2-2ab\cos C\)
Formula-booklet tools for a general triangle. Use the sine rule when you know an angle and its opposite side; use the cosine rule when you know three sides, or two sides and the included angle.
Need the full syllabus wording, formula-booklet reference table, or GDC guidance? See Triangles (Sine/Cosine Rules).
Worked examples
From the top of a cliff 80 m high, the angle of depression to a boat at sea is \(22^\circ\). The boat then moves directly towards the cliff. After moving, the angle of depression is \(47^\circ\).
Find the distance the boat travelled.
Worked solution
Substitute \(h=80\) and \(22^\circ\) into \(\tan\) to find the initial distance from the cliff: \(d_1=\dfrac{80}{\tan22^\circ}\) M1
\(d_1\approx198.1\) m. A1
Substitute \(h=80\) and \(47^\circ\) into \(\tan\) to find the final distance, then subtract: \(d_2=\dfrac{80}{\tan47^\circ}\approx74.6\) m, distance moved \(=198.1-74.6\) M1
\(\approx123\) m. A1
A ladder of length 6 m leans against a wall. The foot of the ladder is on horizontal ground. The wall makes an angle of \(82^\circ\) with the ground (not quite vertical). The ladder makes an angle of \(65^\circ\) with the ground.
(a) Find the angle between the ladder and the wall.
(b) Find how far up the wall the ladder reaches (measured along the wall).
Worked solution
(a) Use the angle sum of a triangle (\(180^\circ\) minus the two known angles): \(180^\circ-65^\circ-82^\circ\) M1
\(=33^\circ.\) A1
(b) Set up the sine rule with 6 m opposite the \(82^\circ\) angle: \(\dfrac{d}{\sin65^\circ}=\dfrac{6}{\sin82^\circ}\Rightarrow d=\dfrac{6\sin65^\circ}{\sin82^\circ}\) M1
\(\approx5.49\) m. A1
Common mistakes
- Reaching for the sine or cosine rule when a right angle already does the job. If the diagram gives you a right angle, plain trigonometry or Pythagoras is usually faster and less error-prone than the general triangle rules.
- Missing that an angle of depression equals the angle of elevation. The angle measured down from the horizontal at the cliff top and the angle measured up from the horizontal at the boat are alternate angles - equal, but easy to mix up under pressure.
- Forgetting to sanity-check the angle sum. A triangle's angles must add to \(180^\circ\) - if your three angles don't, a mistake happened earlier (often a wrong degree/radian mode on the GDC).
Ready to practise properly?
14 mixed-triangle questions, marked instantly like the real exam.
Quick answers
How do I know which triangle tool a question wants?
If there's a right angle, use SOH CAH TOA or Pythagoras. If not, and you know two sides plus a non-included angle, or all three sides, use the sine or cosine rule.
What makes a triangle question "mixed"?
It combines more than one technique in a single problem - often a right-angle triangle for one stage and the sine or cosine rule for another, sometimes wrapped in context like bearings, angles of elevation, or 3D solids.