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