Math, asked by lakshyavalirama1325, 1 year ago

in triangle ac angle ACB = angle CDA. AC=8 cm and AD=3 cm find BD

Answers

Answered by amitnrw
46

Answer:

BD = 18.33 cm

Step-by-step explanation:

in triangle ac angle ACB = angle CDA. AC=8 cm and AD=3 cm find BD

∠ACB  = x

∠CDA = x

CD intersecsect AB at D and Let say

∠DCB = a  then  ∠ACD = x -a

∠CDA = ∠DCB  + ∠CBD

=> x = a + ∠CBD

=> ∠CBD = x - a    (∠CBD = ∠CBA   as D is a point on line BA)

Now in Triangle Δ ABC

AC/ Sin(∠CBA)   = AB/Sin(∠ACB)

=> 8/ Sin(x-a) = AB/Sinx

=> Sin(x)/Sin(x-a) = AB/8    - eq 1

Now in Triangle Δ ACD

AC/Sin(∠CDA)  = AD / Sin(∠ACD)

=> 8/Sin(x)  = 3/Sin(x-a)

=> 8/3 = Sin(x) / Sin(x-a)

=> Sin(x) / Sin(x-a) = 8/3    - eq 2

equating eq 1 & eq 2

AB/8 = 8/3

=> AB = 64/3

AB = AD + BD

=> 64/3 = 3 + BD

=> BD = 64/3 - 3

=> BD = 55/3

=> BD = 18.33 cm

Answered by akshatshukla0987
6

Answer:

Given: ∠ACB=∠CDA

AC=8cm

AD=3cm

In △ABC,∠CDA=90

[given]

∴∠ACB=90

o

From △ACD,cosA=

AC

AD

=

8

3

But from △ABC,cosA=

AB

AC

8

3

=

AB

3

⇒AB=8cm

Then BD=AB−AD

=8−3

=5cm

Please mark my answer as a brainlist

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