In triangle $ABC$ the sides $BC$ , $CA$ and $AB$ are $a$ , $b$ and $c$ respectively, and $\cot C = \dfrac{a}{c}\operatorname{cosec} B - \cot B$ . Given $a = 14.7$ cm, $c = 17.3$ cm and $B = 64.2°$ , find angle $C$ correct to the nearest $0.1°$ .

24.7°

41.5°

50.5°

65.3°

Explanation:

\cot C = \tfrac{14.7}{17.3}\operatorname{cosec}(64.2°) - \cot(64.2°) = 0.9434\times1.1108 - 0.4834 = 0.4604

, so

\tan C = 2.172

and

C = 65.3°

Derived from ZIMSEC Mathematics Paper 1, June 2011, Q6

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