Problems with Triangles
Real triangle problems rarely stop at one formula — you'll often chain the sine rule, cosine rule and area formula together, and watch for the one trap that catches almost everyone: the ambiguous case of the sine rule.
Choosing the right tool
Before combining formulas, make sure the choice between them is automatic. Look at what the question hands you:
Multi-step problems just chain these: use one tool to unlock a missing piece, then feed it into the next. The one genuine trap along the way is the sine rule's ambiguous case — it deserves its own careful treatment before you meet it inside a longer problem.
The ambiguous case of the sine rule
Whenever you use the sine rule to find an angle (rather than a side), there's a subtlety that the cosine rule never has. It comes from a basic fact about sine itself.
$\sin\theta$ is positive for angles in both the first and second quadrant: $\sin\theta = \sin(180° - \theta)$. So if $\sin\theta = 0.7$, your calculator's $\sin^{-1}(0.7)$ gives one angle — but $180°$ minus that angle has exactly the same sine, and is an equally valid solution to the equation. A triangle can genuinely have an obtuse angle whose sine looks identical to an acute one.
This matters specifically when you're given two sides and a non-included angle (the SSA case) and asked to find another angle. The cosine rule never has this problem, because $\cos\theta$ is negative for obtuse angles and positive for acute ones — its inverse always returns a unique angle between $0°$ and $180°$. Sine can't distinguish the two.
Find the principal value $\theta_1 = \sin^{-1}(k)$ first, then test the second candidate $\theta_2 = 180° - \theta_1$. A candidate is valid only if adding it to the other known angle of the triangle still leaves less than $180°$ in total. If it does, both triangles are genuinely possible; if it pushes the angle sum over $180°$, that candidate is impossible and only the acute solution survives. A given angle that's already obtuse or reasonably large will usually rule the second case out immediately.
Worked example 1
Because $a < b$ here, angle $A$ is smaller than angle $B$ — consistent with both $56.4°$ and $123.6°$ being bigger than $30°$. When the side opposite the known angle is shorter than the other given side, watch for genuine ambiguity; when it's longer, the second case almost always fails the angle-sum test.
Worked example 2
A triangle can only have one obtuse angle. Since $A$ is already obtuse at $110°$, $B$ cannot possibly be obtuse too — so the moment you're given (or find) one obtuse angle, any obtuse candidate for a second angle is automatically thrown out without even needing the arithmetic check. Spotting this early saves time.
Worked example 3
The area formula needs two sides and the included angle, but only one side is given here — so the plan is to unlock a second side with the sine rule first, then finish with $\tfrac12 ab\sin C$. This "sine rule, then area" chain is one of the most common multi-step structures in exams.
The subtle step is 4: the area formula needs the angle between the two sides you're using. Here the two sides are $PQ$ and $QR$, which meet at $Q$ — so it's $\sin 70°$, not $\sin 60°$. Always identify which vertex your two chosen sides meet at, and use that angle.
Check yourself
In triangle $ABC$, angle $B = 25°$, $b = 5\,\text{cm}$ and $c = 9\,\text{cm}$. Find the possible value(s) of angle $C$, to 1 d.p.
(Hint: find both candidates from sin C, then test each against the given angle B.)