For the case k = 4 , find the two possible values of QR.

Determine the values of k for which the conditions above define a unique triangle.

attempt to apply cosine rule     M1

\({4^2} = {6^2} + {\text{Q}}{{\text{R}}^2} - 2 \cdot {\text{QR}} \cdot 6\cos 30^\circ\) ( or \({\text{Q}}{{\text{R}}^2} - 6\sqrt 3 {\text{ QR}} + 20 = 0\) )     A1

\({\text{QR}} = 3\sqrt 3  + \sqrt 7 {\text{ or QR}} = 3\sqrt 3  - \sqrt 7 \)     A1A1

[4 marks]

METHOD 1

\(k \geqslant 6\)     A1

\(k = 6\sin 30^\circ  = 3\)     M1A1

Note: The M1 in (b) is for recognizing the right-angled triangle case.

METHOD 2

\(k \geqslant 6\)     A1

use of discriminant: \(108 - 4(36 - {k^2}) = 0\)     M1

k = 3     A1

Note: k = ±3 is M1A0.

[3 marks]

Candidates using the sine rule here made little or no progress. With the cosine rule, the two values are obtained quite quickly, which was the case for a majority of candidates. A small number were able to write down the correct quadratic equation to be solved, but then made arithmetical errors en route to their final solution(s). Part b) was often left blank. The better candidates were able to deduce k = 3 , though the solution \(k \geqslant 6\) was rarely, if at all, seen by examiners.

Candidates using the sine rule here made little or no progress. With the cosine rule, the two values are obtained quite quickly, which was the case for a majority of candidates. A small number were able to write down the correct quadratic equation to be solved, but then made arithmetical errors en route to their final solution(s). Part b) was often left blank. The better candidates were able to deduce k = 3 , though the solution \(k \geqslant 6\) was rarely, if at all, seen by examiners.