In Fig. 6.2, POQ is a line. The value of x is
It is given to us that POQ is a line.
From the figure, we can say that OA and OB are two lines intersecting on the line POQ.
We know,
If a ray stands on a line, then the sum of the two adjacent angles so formed is 180°. - - - - (i)
So, ∠POA + ∠AOB + ∠BOQ = 180°
⇒ 40° + 4x + 3x = 180° (Given - ∠POA = 40°, ∠AOB = 4x, ∠BOQ = 3x)
⇒ 7x = 140°
⇒ x = 20°
Thus, option (A) is the right answer.
Option (B) is not correct because on putting x = 25° on ∠AOB = 4x and on ∠BOQ = 3x, the theorem (i) is not satisfied, i.e.,
∠POA + ∠AOB + ∠BOQ ≠180°
Thus, the value of x is not 25°.
Option (C) is not correct because on putting x = 30° on ∠AOB = 4x and on ∠BOQ = 3x, the theorem (i) is not satisfied, i.e.,
∠POA + ∠AOB + ∠BOQ ≠180°
Thus, the value of x is not 30°.
Option (D) is not correct because on putting x = 25° on ∠AOB = 4x and on ∠BOQ = 3x, the theorem (i) is not satisfied, i.e.,
∠POA + ∠AOB + ∠BOQ ≠180°
Thus, the value of x is not 25°.