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°.

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