answer 10 and 11 questions.
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Q10. X=115
Y=115
Z=35
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10) y+angle(edc) =180
(angle)edc=55. : alternative interior angles
hence
y+ 55=180
y=125
x=y. : angle on same side of transversal
x=125
z+55=90
z=35
11) PAQ +POQ= 180 (opposite angles of cyclic quadrilateral sums 180)
PAQ + 140= 180
PAQ. = 40
(angle)edc=55. : alternative interior angles
hence
y+ 55=180
y=125
x=y. : angle on same side of transversal
x=125
z+55=90
z=35
11) PAQ +POQ= 180 (opposite angles of cyclic quadrilateral sums 180)
PAQ + 140= 180
PAQ. = 40
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