In triangle PQR, q = 8 cm, r = 6 cm and cos P = \(\frac{1}{12}\). Calculate the value of p.
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Correct Answer: Option C
Explanation:
Using the cosine rule, we have
\(p^{2} = q^{2} + r^{2} - 2qr \cos P\)
\(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\)
= \(64 + 36 - 8\)
\(p^{2} = 92 \therefore p = \sqrt{92} cm\)
Using the cosine rule, we have
\(p^{2} = q^{2} + r^{2} - 2qr \cos P\)
\(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\)
= \(64 + 36 - 8\)
\(p^{2} = 92 \therefore p = \sqrt{92} cm\)