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By exterior angle property,
Each of its exterior angles= 360/5= 72 degree
Now, QPU= 72 degree ( exterior angle)
So, since QP is parallel to UV,
QPU + y = 180
➡️ 72 + y = 180
➡️ y= 108 degree
Now, we find x
Interior angle of regular pentagon = 108 degree ( since exterior is 72 )
TQR= 54 degree ( half of interior angle )
Similarly
TRQ= 54 degree
Now, By ASP of a triangle,
x + 54 + 54= 180
➡️ x + 108 = 180
➡️ x = 72 degree
Now, x + y = 72 + 108 = 180 degree
Therefore the answer is
D) 180 degree
Each of its exterior angles= 360/5= 72 degree
Now, QPU= 72 degree ( exterior angle)
So, since QP is parallel to UV,
QPU + y = 180
➡️ 72 + y = 180
➡️ y= 108 degree
Now, we find x
Interior angle of regular pentagon = 108 degree ( since exterior is 72 )
TQR= 54 degree ( half of interior angle )
Similarly
TRQ= 54 degree
Now, By ASP of a triangle,
x + 54 + 54= 180
➡️ x + 108 = 180
➡️ x = 72 degree
Now, x + y = 72 + 108 = 180 degree
Therefore the answer is
D) 180 degree
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