Question

in a triangle ABC AB=52 BC=56 and CA=60. Let D be the foot of the altitude from A and E be the intersection of the internal angle bisector of angle BAC with BC find the length DE​

Answer 1

Answer:

Step-by-step explanation:

=> In ΔABC, as per the cosine rule,

$cos C = a^2 + b^2 - c^2 / 2ab$

$= \frac{56^2 + 60^2 - 52^2}{2*56*60}$

$= \frac{3136+3600-2704}{2*56*60}$

$= \frac{4032}{2*56*60}$

= 0.6

=> In ΔADC, cosC = DC/AC

DC = AC CosC

= 60 * 0.6

=36

=> As AE is angle bisector,

$\frac{AB}{BE} = \frac{AC}{EC}$

$\frac{AB}{AC} = \frac{BE}{EC}$

∴$\frac{BE}{EC} = \frac{52}{60}$

∴$\frac{BE+EC}{EC} = \frac{52+60}{60}$  (compenendo)

$\frac{BC}{EC} = \frac{112}{60}$

∴ EC = BC * $\frac{60}{112} = 56 * \frac{60}{112}$

∴ EC = 30

but, DE = DC - EC

∴ DE = 36 - 30 = 6

Thus, the length of DE is 6