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# Tangents of circles problem (example 2)

Sal finds a missing angle using the property that tangents are perpendicular to the radius. Created by Sal Khan.

## Want to join the conversation?

• At , why is a line that is tangent to a circle perpendicular to the radius of that circle?
• It is because any line that is tangent to a circle only touches one point on the circle and a line from any point on the circle to the center of the circle is a radius. Therefore, any line that is tangent to the circle would also be perpendicular to a radius.
• What is tangent? At .
• A tangent is a line on the outside of a circle or curve that only touches at one point.
• Do opposite angles in a quadrilateral always supplemental (add up to 180 degrees)?
• In general, the answer is no. However, all cyclic quadrilaterals ( quadrilaterals that can have a circumscribed circle drawn) do have opposite angles that add up to 180 degrees. It is a special property of those quadrilaterals, and it can be used to prove that a quadrilateral is cyclic.
• What is a cyclic quadrilateral?
• A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. In other words, it is a quadrilateral that has a circumcircle. A quadrilateral is cyclic if and only if the opposite angles are supplementary.
• am I the only one or is he speaking really slowly
• If you feel that way , you can change the listening speed to a higher rate!
• For any given arc on the circle, will there be only one possible circumscribed angle?
• At , Sal said the sides are going to add up to 360 degrees, but meant its angles.
• Correct he did mean that, nice catch. There is also a little note box correcting him too at the bottom right.
• Is the circumscribed angle always double the inscribed angle
• Why is an inscribed angle double the central angle? I don't get what the arc has to do with it either.
(1 vote)
• The three points (𝐵, 𝐶, 𝐷) that form the inscribed angle also form two isosceles triangles together with the circle center 𝑂,
namely △𝐵𝑂𝐷 and △𝐶𝑂𝐷.

Thereby, 𝑚∠𝑂𝐵𝐷 + 𝑚∠𝑂𝐶𝐷 = 𝑚∠𝐵𝐷𝐶

Thus the measures of three of the four interior angles of quadrilateral 𝑂𝐵𝐶𝐷

The fourth interior angle is 360° − 𝑚∠𝐵𝑂𝐶

The sum of all four angles add to 360°, which gives us
2⋅𝑚∠𝐵𝐷𝐶 + 360° − 𝑚∠𝐵𝑂𝐶 = 360°
⇒ 𝑚∠𝐵𝐷𝐶 = (1∕2)⋅𝑚∠𝐵𝑂𝐶

In other words, the measure of the inscribed angle is half the measure of the central angle.