question_answer

In any triangle \[AB=2,BC=4,CA=3\]and D is mid point of BC, then [Roorkee 1995]

A) \[\cos B=\frac{11}{6}\]

B) \[\cos B=\frac{7}{8}\]

C) \[AD=2.4\]

D) \[A{{D}^{2}}=2.5\]

Correct Answer: D

Solution :

\[\cos B=\frac{{{2}^{2}}+{{4}^{2}}-{{3}^{2}}}{2\times 2\times 4}=\frac{11}{16}\] \ \[\frac{11}{16}=\frac{{{2}^{2}}+{{2}^{2}}-A{{D}^{2}}}{2\times 2\times 2}\Rightarrow A{{D}^{2}}=2.5\].