In triangle PQR, q = 8 cm, r = 6 cm and cos P = \(\frac{1}{12}\). Calculate the value of p.
- \(\sqrt{108}\) cm
- 9 cm
-
\(\sqrt{92}\) cm
- 10 cm
The correct answer is: C
Explanation
Using the cosine rule, we have
\(p^{2} = q^{2} + r^{2} - 2qr \cos P\)
\(p^{2} = 8^{2} + 6^{2} - 2(8)(6)(\frac{1}{12})\)
= \(64 + 36 - 8\)
\(p^{2} = 92 \therefore p = \sqrt{92} cm\)
There is an explanation video available .