In $ \triangle \mathrm{ABC} $, the bisector of $ \angle \mathrm{A} $ intersects $ \mathrm{BC} $ at $ D $. If $ A C=4.2, D C=3 $ and $ B C=5 $, find $ \mathrm{AB} $.
Given:
In \( \triangle \mathrm{ABC} \), the bisector of \( \angle \mathrm{A} \) intersects \( \mathrm{BC} \) at \( D \).
\( A C=4.2, D C=3 \) and \( B C=5 \).
To do:
We have to find \( AB \).
Solution:
We know that,
An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.
Therefore,
$\frac{AB}{AC}=\frac{BD}{DC}$
$\frac{AB}{4.2}=\frac{BC-DC}{3}$
$\frac{AB}{4.2}=\frac{5-3}{3}$
$\frac{AB}{4.2}=\frac{2}{3}$
$AB=\frac{4.2\times2}{3}$
$AB=2.8\ cm$
Hence, the value of $AB$ is $2.8\ cm$.
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