In triangle PQR, \(q = 8\text{ cm}\), \(r = 6\text{ cm}\) and \(\cos P = \frac{1}{12}\). Calculate the value of \(p\).

Answer Details

Using the cosine rule, we have

p2=q2+r2−2qrcosP $p^2=q^2+r^2-2qr\cos P$

p2=82+62−2(8)(6)(112) $p^2=8^2+6^2-2(8)(6)(\frac{1}{12})$

= 64+36−8 $64+36-8$

p2=92∴p=92−−√cm