In a parallelogram ABCD, AP and CQ are drawn perpendicular to the diogonal BD on measuring it is found that angle PAB =65 and angleDAB =75 find the measure of angle QCD

Dear Student,

Please find below the solution to the asked query:

We form our diagram , As :



Now from angle sum property of triangle we get in triangle ABP :

$\angle$ PAB + $\angle$ APB + $\angle$ ABP = 180$°$ , NOw we substitute values from above diagram we get :

65$°$  + 90$°$ + $\angle$ ABP = 180$°$ ,

$\angle$ ABP = 25$°$                                                        --- ( 1 )

And

$\angle$ ABP = $\angle$ CDQ                                              ( Alternate interior angles as AB | | CD and BD is transversal line )

From equation 1 we get :

$\angle$ CDQ = 25$°$                                                        --- ( 2 )

Now from angle sum property of triangle we get in triangle CDQ :

$\angle$ CDQ + $\angle$ CQD + $\angle$ QCD= 180$°$ , Now we substitute values from above diagram and from equation 2 we get :

25$°$  + 90$°$ + $\angle$ QCD = 180$°$ ,

$\angle$ QCD = 65$°$                                                                       ( Ans )

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards