Question

Find x . If u are a mathematician. Pls do . I need it

Answer 1

Answer:

Here OB and CD are 2 intersecting chords

As 2 intersecting chords always make same corresponding angle to chord or lines joining a point other than respective intersecting chord

So, <COB = <CDB =x

As <COA =130

SO, <COB = 180-130( linear pair)

= 50

Hence , <CDB =X = 50

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Answer 2

❏ Solution:-

•Fig:-

is in the attachment (of the questionar)

•Given:-

$\sf\longrightarrow \angle AOC=130\degree$

$\sf\longrightarrow OC||BD$

•To Find:- $\sf\longrightarrow \angle BDC=X=?$

•construction:-

O,D and B,C are connected.

•Finding:-

Now, $\sf\longrightarrow \angle AOC=130\degree$

$\sf\longrightarrow \angle COB=\angle AOB-\angle AOC$

$\sf\longrightarrow \angle COB=180\degree-130\degree$

$\sf\longrightarrow\boxed{\red{ \angle COB=50\degree}}$

Now, we know that radius of a circle bisects the chord.

Here, radius OB , bisected the chord CD.

$\therefore \angle COD$ is bisected by OB.

$\therefore \angle DOB=\angle COB=50\degree$

Hence,

$\sf\longrightarrow \angle COD=\angle DOB+\angle COB$

$\sf\longrightarrow \angle COD=50\degree+50\degree$

$\sf\longrightarrow\boxed{ \angle COD=100\degree}$

Now, OC=OD (as, radii of same circle)

∴ $\therefore \angle OCD=\angle ODC$

Now, from the ∆COD,

$\sf\longrightarrow \angle COD+\angle OCD+\angle ODC=180\degree$

$\sf\longrightarrow 100\degree+\angle OCD+\angle OCD=180\degree$

$\sf\longrightarrow 2\angle OCD=180\degree-100\degree$

$\sf\longrightarrow \angle OCD=\frac{\cancel{80}\degree}{\cancel2}$

$\sf\longrightarrow \bf\angle OCD= 40\degree$

Now, OC||BD,

So, $\sf\longrightarrow \bf\angle OCD= \angle BDC=40\degree$

[as, alternate angle]

$\sf\longrightarrow \bf\boxed{ \red{\angle BDC=X=40\degree}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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