Extend CDO' to cut the circle O' at E. Join AE, BE. Let angle AEB = y By the result of Problem 717, y = 65 deg But AEBD is a cyclic quadrilateral. So x + y = 180 deg Hence x = 115 deg
x = angle subtended by major arc AB of circle(O')at D = half the ∠ subtended by arc AB of circle (O')at center O'. = (1/2)major∠AO'B = (1/2)(360°-minor∠AO'B) = (1/2)[360°-(180°-∠ACB)] (∵ AO'BC is a cyclic quadrilateral) = (1/2)(180°+∠ACB) = (1/2)(180°+50°) = (1/2)(230°) = 115°
Extend CDO' to cut the circle O' at E.
ReplyDeleteJoin AE, BE. Let angle AEB = y
By the result of Problem 717,
y = 65 deg
But AEBD is a cyclic quadrilateral.
So x + y = 180 deg
Hence x = 115 deg
By applying the formula that angle bda=90+bca/2 therefore half of 50=25 and by adding 90 we get 90+25=115
ReplyDeletetherefore measure of angle bda =115
x
ReplyDelete= angle subtended by major arc AB of circle(O')at D
= half the ∠ subtended by arc AB of circle (O')at center O'.
= (1/2)major∠AO'B
= (1/2)(360°-minor∠AO'B)
= (1/2)[360°-(180°-∠ACB)] (∵ AO'BC is a cyclic quadrilateral)
= (1/2)(180°+∠ACB)
= (1/2)(180°+50°)
= (1/2)(230°)
= 115°
<AO'B=180-50=130
ReplyDelete<ADB=180-<AO'B/2=180-65=115