Question

plsss...some1 answer my question plsssssss

Answer 1

ang.BOC =Y
ANG. OBA AND ANG.OAB=2Y(EXTERIOR ANGLE THEOREM)
SO ANG.AOB=180-4Y
x+(180-4y)+y=180
x+180-3y=180
x-3y=0
x=3y

Answer 2

Answer:  Proved.

Step-by-step explanation:  In the given figure, 'O' is the centre of the circle and BC = OB.

We are given to prove that x = 3y.

In ΔOBC, we have

OB = BC,

therefore ∠OCB = ∠BOC = y (Angles opposite to equal sides are equal).

Since the sum of two remote interior angles in a triangle is equal to the exterior angle, so we have

$\angle ABO=\angle OCB+\angle BOC\\\\\Rightarrow \angle ABO=y+y\\\\\Rightarrow \angle ABO=2y.$

Now, in ΔAOB, OA = OB =radius of the circle.

Therefore,

∠ABO = ∠BAO (Angles opposite to equal sides are equal).

So,

∠ABO = ∠BAO = 2y.

Again, we have from ΔAOB that

$\angle AOB+\angle ABO+\angle BAO=180^\circ\\\\\Rightarrow \angle AOB+2y+2y=180^\circ\\\\\Rightarrow AOB=180^\circ-4y.$

Now, we have

$\angle DOA+\angle AOB+\angle BOC=180^\circ~~~~\textup{(straight angle)}\\\\\Rightarrow x+(180^\circ-4y)+y=180^\circ\\\\\Rightarrow x-3y=0\\\\\Rightarrow x=3y.$

Hence proved.