Question

pls solve it fast

u have to find ab and bc​

Answer 1

Answer:

in triangle ABC,

angle A is 30..

let ab =x and bc=y

tan A=P/B

tan30=y/x

1/√3=y/x

we get ,x=√3y. [equation 1]

In triangle ADB,

angle A=60

tan 60=DB/AB

√3=(20+y)/x

from here we get,x=(20+y)/√3. [equation.2]

from equation 1 and equation 2 we get,

(20+y)/√3 =√3y

20+y=3y

20=2y

y=10cm

from equation 1 we have,

x=√3y

so,x=10√3cm

ab=10√3cm

bc=10cm

hope it helps u ☺️☺️☺️

Answer 2

Answer:

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Step-by-step explanation:

$Given : CD = 20cm$

such that /_ ACB = 60°

and, /_ADB = 30°

$To \: Find : AB \: and, \: BC$

$In \: r.t \: \: ∆ABD, we \: have:$

$\frac{AB}{BD} = Tan \: 30°$

$=> \frac{AB}{20+x} = \: \frac{1}{√3}$

$=>√3AB = 20+x$

$=> AB = \frac{20+x}{ \sqrt{3} } ---(1)$

$Now, \: In \: r.t. \: \: ∆ ABC , we \: have :$

$\frac{AB}{BC} = Tan \: 60°$

$=>\frac{20+x}{ \frac{ \sqrt{3} }{x} } = √3$

$=> \frac{20+x}{ \sqrt{3x} } = \frac{ \sqrt{3} }{1}$

$=> 20+x = 3x$

$=> 20= 3x-x$

$=> 2x = 20$

$=> x = \frac{20}{2} =10$

$Now , substituting \: the \\ \: value \: of \: x \: in \: (1) , \: we \: get :$

$AB = \frac{ 20+10}{ \sqrt{3} }$

$=> \frac{30}{ \sqrt{3} } × \frac{ \sqrt{3} }{ \sqrt{3} }$

$=> \frac{30√3}{3}$

$=> 10√3$

$= 10×1.73$

$= 17.3$

$Therefore, AB = 17.3m \\ \: and, \: \: \: BC = 10m.$

$please \: like \: my \: answer \\ \&\\ mark \: me \: brainliest \: please \: :)))$

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