Given: Δ ABC and AD bisects ∠A, meeting side BC at D. And AC = 4.2 cm, DC = 6 cm, and BC = 10 cm.
Required to find: AB
Since, AD is the bisector of ∠A meeting side BC at D in Δ ABC
⇒ \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\)
\(\frac{AB}{4.2}\) = \(\frac{BD}{6}\)
We know that,
BD = BC – DC
= 10 – 6
= 4 cm
⇒ \(\frac{AB}{4.2}\) = \(\frac{4}{6}\)
AB = \(\frac{(2 \times 4.2)}{3}\)
∴ AB = 2.8 cm