Find the angles of a quadrilateral ABCD inscribed in a circle if the angle CBD = 48, angle ACD = 34, angle BDC = 64.

| November 25, 2020 | education

Consider a triangle СОD
O-point of intersection of diagonals
Sum of angles 180
OCD angle + OBC angle + COD angle = 180
Angle COD = 180- (34 + 64) = 82
BOC angle + COD angle = 180 as adjacent
BOC angle = 180-82 = 98
Consider the BOS triangle
Angle BCO = 180-Angle CBO-angle BOC
Angle ВСО = 180-48-98 = 34
Angle С = angle ВСО + angle ОSD = 34 + 34 = 68
The sum of the opposite angles of the inscribed quadrilateral is 180
Angle C + Angle A = 180
Angle A = 180-68 = 112
Consider a triangle ABD
ABD + ADB = 180-112 = 68
Let ABD = x
ADB = y
X = 68-y
Consider a triangle OAD
OAD = 180-98-68-y
OAD = 82-68-y = 14-y
Consider the triangle OAB
OAB = 180-y-82 = 98-y
98th + 14th = 68
2y = 98 + 14-68
2y = 16
y = 8
x = 68-8
x = 60
Angle B = ABD + CBD = 60 + 48 = 108
Angle D = OAD + BDC = 8 + 64 = 72
Answer: angle A = 112
Angle B = 108
Angle C = 68
Angle D = 72



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