Dear Student, sinB=45,tanC=1,BD=15 cm cosB=25-165=35AD=DCtan2B-1cos2B=-1LHS=sin2Bcos2B-1cos2B=452-1352=-99=-1hence proved Regards

Please solve 8c(ii)

Please solve 8c(ii) AD BC. D is of W, SO BD In figure AABC is with AB AC sin C tan C — cot B. (2) at B. Givm that ZAC-B Side AB 2 and side BC t unit. find value of (3' AD is to BD • 15 (i) calculate the lengths of AD, AB, DC and AC. (2) (a) Draw AD perpendicular to BC, then BD DC = 3 cm and AD 4 cm.

Dear Student,

\sin B = \frac{4}{5}, \tan C = 1, B D = 15 c m \cos B = \frac{\sqrt{25 - 16}}{5} = \frac{3}{5} A D = D C \tan^{2} B - \frac{1}{\cos^{2} B} = - 1 L H S = \frac{\sin^{2} B}{\cos^{2} B} - \frac{1}{\cos^{2} B} = \frac{{(\frac{4}{5})}^{2} - 1}{{(\frac{3}{5})}^{2}} = \frac{- 9}{9} = - 1 h e n c e p r o v e d . R e g a r d s