Question

plz answer this question5

Answer 1

Hey Mate !

Here is your solution :

Extend AC to E on QB.

Now,

=> <CEB + <PAC = 180° ( Angles on the same side of transversal )

=> <PAC = 180° - <CEB ------- ( 1 )

And,

=> <ECB + <BCA = 180° (Linear Pair of angles )

=> <BCA = 180° - <ECB ------ ( 2 )

And,

=> <CBE + <CBQ = 180° ( Linear Pair of angles )

=> <CBQ = 180° - <CBE ----- ( 3 )

In ∆CEB,

=> <CBE + <CEB + <ECB = 180° ---- ( 4 )

By adding ( 1 ) , ( 2 ) and ( 3 ),

=> <PAC + <BCA + <CBQ = 180° - <CEB + 180° - <ECB + 180° - <CBE

=> <PAC + <BCA + <CBQ = 540° - ( <CEB + <ECB + <CBE )

By substituting the value of ( 4 ),

=> <PAC + <BCA + <CBQ = 540° - 180°


=> <PAC + <BCA + <CBQ = 360°


The required answer is 360°.

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Hope it helps !! ^_^