Question

In the given figure qr=qt,sp=st,ts=qt.prove that angle r =3*p

Answer 1

Answer:

Given : QR  =  QT , so from base angle theorem we get in Δ∆ QRT :

∠∠ R = ∠∠ QTR                                              --- ( 1 )

And given : SP  =  ST , So from base angle theorem we get in Δ∆ PST :

∠∠ P = ∠∠ STP                                              --- ( 2 )

And from angle sum property of triangle we get in Δ∆ PST :

∠∠ P  + ∠∠ STP +  ∠∠ PST = 180°° , Substitute value from equation 2 we get :

∠∠ P  + ∠∠ P +  ∠∠ PST = 180°° ,  

∠∠ PST = 180°° - 2 ∠∠ P                                  --- ( 3 )

And

∠∠ QST + ∠∠ PST  = 180°°                       ( Linear pair angles )

∠∠ QST + 180°° -  2 ∠∠ P  = 180°°           ( From equation 3 )

∠∠ QST  =  2 ∠∠ P                                             --- ( 4 )

And given : TS  =  QT , So from base angle theorem we get in Δ∆ QST :

∠∠ SQT = ∠∠ QST ,

∠∠ SQT =  2 ∠∠ P                                    ( From equation 4 )

And from angle sum property of triangle we get in Δ∆ QST :

∠∠ QST  + ∠∠ SQT +  ∠∠ QTS = 180°° , Substitute value from above equation and from equation 4 we get :

2 ∠∠ P  + 2 ∠∠ P +  ∠∠ QTS = 180°° ,  

∠∠ QTS = 180°° - 4 ∠∠ P                                  --- ( 5 )

And

∠∠ STP + ∠∠ STR  =  180°°                      ( Linear pair angles )

∠∠ STP + ∠∠ QTS + ∠∠ QTR  =  180°°     ( From given diagram : ∠∠ STR = ∠∠ QTS + ∠∠ QTR )

Now we substitute values from equation 1 , 2 and 5 and get :

∠∠ P  + 180°° - 4 ∠∠ P + ∠∠ R = 180°°

∠∠ R  - 3 ∠∠ P = 0 ,

∠∠ R  = 3 ∠∠ P                                           ( Hence proved )

Step-by-step explanation:

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