10. Shown is a circle centre o
BC and D are points on the circumference
ABO S a straight line
AC is a tangent to the circle.
320
A. Find the mZAOC
B. Find the mZBDC
C. Find the mZACD
Answers
AB is a a straight line and AC is a tangent to the circle, then the
(a) ∠AOC = 58°
(b) ∠BDC = 32°
(c) ∠ACD = 148°
According to the figure,
A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted \triangle ABC. In Euclidean geometry, any three points, when non-collinear, determine a unique triangle and simultaneously, a unique plane.
What is the tangent of the circle?It is the line which intersects the circle exactly at one point.
(a) Find m ∠AOC
Sum of three angles of a triangle is = 180°
So, in Triangle AOC = 180°
∠A + ∠O + ∠C = 180°
See in Image, ∠A and ∠C are 90° and 32°
∠A + ∠O + ∠C = 180°
We can substitute these values,
∠O = 180 - ∠A - ∠C
∠O = 180 - 90 - 32
∠O = 58°
∠AOC = 58°
(b) ∠BDC is a parallelogram,
So,
Opposite angle is equal,
∠BDC = ∠ABC
∠BDC = 32° ( since ∠ABC = 32° )
(c) Find m ∠ACD
Sum of three angles of a triangle is = 180°
So, in Triangle ACD = 180°
∠A + ∠C + ∠D = 180°
See in Image, ∠A and ∠D are 90° and 32°
∠A + ∠D + ∠C = 180°
We can substitute these values,
∠C = 180 - ∠A - ∠D
∠C = 180 - 90 - 32
∠C = 58°
Now,
∠ACD = 90 + 58
∠ACD = 148°
Therefore,
The circumference of tangent to the circle,
(a) ∠AOC = 58°
(b) ∠BDC = 32°
(c) ∠ACD = 148°
To learn more about information visit Tangent of the circle :
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