Q. 355.0( 2 Votes )

In Fig. 5.28, PQ A. (2 + b)°

B. (3–b)°

C. (108–b)°

D. (180–b)°

Answer :

1 = (2a + b)° …Equation (i)


6 = (3a–b)°


Since 5 and 6 forms a linear pair so


5 = (180-3a + b)°


5 = 1 = (180-3a + b)° (Corresponding angles) …Equation (ii)


Equating Equation (i) and Equation (ii) we get


2a + b = 180-3a + b


5a = 180


a = 360


Since 1 and 2 forms a linear pair so


2 = 1800- 2a-b


2 = (108-b)0

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