SPS SPS FM 2020 December — Question 11 6 marks

Exam BoardSPS
ModuleSPS FM (SPS FM)
Year2020
SessionDecember
Marks6
TopicSine and Cosine Rules
TypeAmbiguous case (two solutions)
DifficultyStandard +0.3 This is a straightforward sine rule application with the ambiguous case. Part (i) requires using the sine rule to find two solutions (a standard textbook exercise), and part (ii) asks for the condition for uniqueness, which is a well-known result (when the given side opposite the known angle equals the perpendicular height, k sin 30° = 6, giving k = 12). Both parts are routine applications of standard A-level content with no novel problem-solving required.
Spec1.05b Sine and cosine rules: including ambiguous case

In the triangle \(PQR\), \(PQ = 6\), \(PR = k\), \(P\hat{Q}R = 30°\).
  1. For the case \(k = 4\), find the two possible values of \(QR\) exactly. [3]
  2. Determine the value(s) of \(k\) for which the conditions above define a unique triangle. [3]

In the triangle $PQR$, $PQ = 6$, $PR = k$, $P\hat{Q}R = 30°$.
\begin{enumerate}[label=(\roman*)]
\item For the case $k = 4$, find the two possible values of $QR$ exactly. [3]
\item Determine the value(s) of $k$ for which the conditions above define a unique triangle. [3]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM 2020 Q11 [6]}}