Question

In any triangle $AB = 2,BC = 4,CA = 3$ and $D$ is mid point of BC, then

• A. $\cos B = \frac{11}{6}$
• B. $\cos B = \frac{7}{8}$
• C. $AD = 2.4$
• D. $AD^{2} = 2.5$

Answer 1

Answer: (D)

$AD^{2} = 2.5$

Answer 2

From $\Delta ABC$, $\cos B = \frac{2^{2} + 4^{2} - 3^{2}}{2 \times 2 \times 4} = \frac{11}{16}$

From $\Delta ABD$, $\cos B = \frac{2^{2} + 2^{2} - AD^{2}}{2 \times 2 \times 2} = \frac{11}{16}$

$\therefore$ $\frac{11}{16} = \frac{2^{2} + 2^{2} - AD^{2}}{2 \times 2 \times 2}$

⇒ $AD^{2} = 2.5$.