JAMB 2013 · UTME · Q33

Sine & cosine rules

In triangle $PQR$ , $q = 8\text{ cm}$ , $r = 6\text{ cm}$ and $\cos P = \frac{1}{12}$ . Find $p$ .

Worked solution (try it first)

$p$ faces angle $P$ , which lies between sides $q$ and $r$ .
Cosine rule: $p^2 = q^2 + r^2 - 2qr\cos P$ .
Put in the numbers: $p^2 = 64 + 36 - 2(8)(6) \times \frac{1}{12}$ .
The last term is $96 \div 12 = 8$ .
So $p^2 = 100 - 8 = 92$ and $p = \sqrt{92}$ cm, option C.

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