a. To prove that ∠BOA = 90°, we need to use the properties of tangents and radii. Since OA and OB both intersect the circle at point O, they are both radii of the circle. This means that OA and OB are congruent, making triangle OAB isosceles. In an isosceles triangle, the base angles are equal. Therefore, ∠OAB = ∠OBA.
Now let's consider ∠BOA. By the angle sum theorem of a triangle, the sum of the angles in a triangle is 180°. Since ∠OAB = ∠OBA, we can say that 2∠OAB + ∠BOA = 180°. This simplifies to 2∠OAB + ∠BOA = 180°. But since ∠OAB = ∠OBA, we can rewrite this as 2∠OBA + ∠BOA = 180°. Since the sum of the angles in triangle OAB is 180°, ∠OBA + ∠OAB + ∠BOA = 180°. Substituting in for ∠OAB = ∠OBA, we get 2∠OBA + ∠BOA = 180°. Simplifying this equation gives us ∠OBA + ∠BOA = 90°. But we know that ∠OBA = ∠OAB, so we can substitute this in: ∠OAB + ∠BOA = 90°. Therefore, ∠BOA = 90°.
b. Since we have already proven that ∠BOA = 90°, and we are given that ∠AOP = 64°, we can find ∠BOD by using the fact that the angle formed between a tangent and a radius is always 90°. This means that ∠BOA = 90°.
Since the sum of angles in a triangle is 180°, we can calculate m∠BOD as follows:
m∠BOD = 180° - 64° - 90°
m∠BOD = 26°
Therefore, the measure of angle BOD is 26°.
In circle O, and are
tangents. The figure is not drawn
to scale.
a. Prove that .
b. Find m BOD for m AOP =
64°. Explain your reasoning.
(4 points)
1 answer
1 answer