﻿ Circles: The Angle formed by a Chord and A Tangent, Intercepted Arc. Examples, Pictures, Interactive Demonstration and Practice Problems Chord, Tangent and the Circle

The Intersection of a Tangent and Chord

An angle formed by a chord (link) and a tangent (link) that intersect on a circle is half the measure of the intercepted arc.

$x = \frac 1 2 \cdot \text{ m } \overparen{ABC}$

Note: Like inscribed angles, when the vertex is on the circle itself, the angle formed is half the measure of the intercepted arc.

Interactive Applets

$$\angle X = \class{data-anglex-0}{35.92} \\ \overparen{\rm ABC} = \class{data-anglex-1}{35.92} \\ \frac 1 2 \overparen{\rm ABC} = \class{data-anglex-0}{35.92}$$

$$\angle Y = \class{data-angley-0}{35.92} \\ \overparen{\rm ADC} = \class{data-angley-1}{35.92} \\ \frac 1 2 \overparen{\rm ADC} = \class{data-angley-0}{35.92}$$

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Practice Problems

Problem 1

Chord $$AC$$ intercepts a tangent tangent at point C. If the measure of $$\overparen{abc} = 70^{\circ}$$, what is x ?

Use the theorem above to find the measure of angle formed by the intersection of the tangent that intersects chord AC.

By the theorem, the measure of angle is half of the intercepted arc which is $$70 ^{\circ}$$ .

Therefore $$x = \frac{1}{2} \cdot 70 = 35^{\circ}$$ .

Problem 2

Chord $$AC$$ intercepts a tangent tangent at point C. If the measure of $$\overparen{abc} = 110^{\circ}$$, what is x ?

What is the $$m\overparen{ABC}$$(the measure of arc ABC) ?

Remember the theorem: the angle formed by the tangent and the chord is half of the measure of the intercepted arc.

Therefore, the arc is double the angle. $$m\overparen{ABC} = 2 \cdot 110^{\circ}=55^{\circ}$$

Challenge Problems

Problem 3

For the $$m\overparen{ABC}$$ ? to equal ¾ the total measure of the circle's circumference, what must be the value of X?

Total measure of circle's circumference = 360°.

$$\frac 3 4 (360^{\circ}) = 270^{\circ}$$

By our theorem, we know that the angle formed by a tangent and a chord must equal half of the intercepted arc so $$x = \frac 1 2 \cdot 270^{\circ} =135^{\circ}$$ .

Problem 4

Look at Circle 1 and Circle 2 below. In only one of the two circles does a tangent intersect with a chord. Which circle is it?

Circle 1 is the only circle whose intercepted arc is half the measure of the angle between the chord and the intersecting line.

Problem 5

What is the measure of $$\angle ACZ$$ for circle with center at O?

The key to this problem is recognizing that $$\overline{AOC}$$ is a diameter.

Therefore, $$m \overparen{ABC} = 180 ^{\circ}$$.

At this point, you can use the formula, $$\\ m \angle ACZ= \frac{1}{2} \cdot 180 ^{\circ} \\ m \angle ACZ= 90 ^{\circ}$$

Problem 6

$$\overparen{MNJ} : \overparen {JLM}$$ is 3:2, what is the $$m\angle MJK$$?

The key to this problem is recognizing that the total degrees in a circle is $$360^{\circ}$$ .

From there you can set up an equation using the 3:2 ratio.

$$3x + 2x =360 \\ 5x = 360 \\ \frac{5x}{5} = \frac{360}{5} \\ x = 72 \\ \overparen {JLM }= 2x = 2 \cdot 72 \\ \overparen {JLM }=144 ^{\circ}$$

At this point, you can use the formula, $$\\ m \angle MJK= \frac{1}{2} \cdot 144 ^{\circ} \\ m \angle MJK = 72 ^{\circ}$$