75 POINTS!!!! Create a trianlge ABC. Choose two of the angles on ∆ABC, and locate the line segment between them. Draw a new line
segment, , parallel to the line segment you located on ∆ABC. You can draw of any length and place it anywhere on the coordinate plane, but not on top of ∆ABC. From points D and E, create an angle of the same size as the angles you chose on ∆ABC. Then draw a ray from D and a ray from E through the angles such that the rays intersect. You should now have two angles that are congruent to the angles you chose on ∆ABC. Label the point of intersection of the two rays F, and draw ∆DEF by creating a polygon through points D, E, and F. Please show me a picture as your answer
2 answers:
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Answer:
30°
Step-by-step explanation:
Call the other end of the chord point B and the center of the circle point O. Then triangle AOB is an equilateral triangle, since OA = OB = AB.
Angle OAB is the internal angle of that triangle, so is 60°. Since OA is perpendicular to the tangent line (makes an angle of 90°), The angle between the tangent line and the chord must be ...
90° - 60° = 30°
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The other way you know this is that central angle AOB is 60°, and the tangent-to-chord angle is half that, or 30°.
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One way to remember the angle relationship between a tangent line and a chord is this:
Consider a point C on long arc AB. The measure of inscribed angle ACB is half the measure of central angle AOB, no matter where C is on the circle. (If C happens to be on the short arc AB, then central angle AOB is a reflex angle, but the relationship still holds.) Consider what happens when C approaches A. The angle at vertex C is still the same: 1/2 the measure of central angle AOB. This remains true even in the limit when points A and C become coincident and line AC is a tangent at point A.
