ab and ac are tangents to a circle with centre o. if angle bac is equal to 40 degree, find angle abc and angle acb.
angles OBA and OCA are right
angle BOC is 90-40 = 50
ABC =180 - 20 - 90 = 70 = ACB
Below is the circle with center o. Given 70⁰ ,find BAC
To find angle ABC and angle ACB, we can use some properties of tangents and angles formed by intersecting lines.
1. We know that AB and AC are tangents to the circle with center O. Therefore, both AB and AC are perpendicular to the radii of the circle drawn from point O.
2. As a result of property 1, triangle BOC is an isosceles triangle because both OB and OC are radii of the same circle.
3. Since triangle BOC is isosceles and angle BOC is a central angle (angle at the center of the circle), we can conclude that angle BOC is equal to angle BCO.
Now, let's solve for the angles:
1. Angle BAC is given as 40 degrees.
2. From property 3, we know that angle BOC is equal to angle BCO. Let's assume it to be x degrees.
3. In triangle ABC, the sum of the angles is 180 degrees.
angle BAC + angle ABC + angle ACB = 180 degrees
40 degrees + angle ABC + angle ACB = 180 degrees
4. Now, using property 2, we know that angle BAC is equal to angle ABC (as they are alternate angles formed by parallel lines AB and OC). Therefore, angle ABC is also equal to 40 degrees.
40 degrees + 40 degrees + angle ACB = 180 degrees
80 degrees + angle ACB = 180 degrees
5. Solving for angle ACB:
angle ACB = 180 degrees - 80 degrees
angle ACB = 100 degrees
So, the angle ABC is 40 degrees, and the angle ACB is 100 degrees.