Question

From the following figure , find the value of x.​

Answer 1

Answer:

Option c is correct...

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Answer 2

⭐Given:

• O is the centre of a circle

• Seg BT perpendicular on seg TN.i.e. <T=90°

• <TNO= 30°

⭐To Find:

• Value of x=?

⭐Solution:

In ∆ NTO <NOB is an exterior angle

: .By exterior angle therom,

<NOB=<OTN+TNO

: .<NOB=90°+30°

: .<NOB=120°.......... (1)

NOW,

In ∆AOB,

OA & OB are radius.

: . OA=OB.............(2)

: . <NOB+<OBA+<OAB=180°.........(Sum of all angles of triangle is 180°)

Here, <OBA≈ <OAB .........(Angles opposite to the congruent sides are Congruent.)

Let, Substitute <OBA=<OAB=x

: . <NOB+x+x=180°

: . 120°+2X=180°

: . 2x=180-120

: . 2x= 60

: .x=60/2

: . x=30°

Therefore,

The value of x is 30 °.