Word Lesson: Circles - Angles from Secants and Tangents
By M Ransom
In order to solve problems which involve secants, tangents, and angles formed by them, it is necessary to
- know the basic properties of the angles and segments formed by secants and tangents,
- know that an angle inscribed in a circle has a measure of ½ the intercepted arc, and
- be able to solve basic linear equations.
A typical problem involving the angles formed by secants and tangents in a circle gives us information about the measures of the angle exterior to the circle and/or about the measures of the intercepted arcs of the circle. Two examples of this type of problem are presented below.
- In circle O shown below, two secants from point P intercept arcs CB = x – 10 and AD = 2x. What is the measure of arc AD if angle P is 25°?

We know that the measure of an external angle P when formed by two secants is equal to one half the difference of the measures of the intercepted arcs.
25 = (1/2)(x + 10)
50 = x + 10
x = 40ºSince we were given that arc AD = 2xAD = 2(40º) = 80°
- In circle O shown below, angle P is x°, arc CB is 55°, and arc CD is 4x – 9. What is the measure of arc CD?

We know that angle P must equal ½ the difference of the measures of arcs CD and CB.
2x = 4x – 64
2x = 64
x = 32Since we were given that arc CD = 4x – 9CD = 4(32) – 9
CD = 119°
When an angle is formed outside a circle either by two secants or one secant and a tangent, there is a relationship between the measure of this angle and the difference of the measures of the intercepted arcs.
angle outside = ½ (difference of intercepted arcs)
We can use this equation to relate measures and in some cases, find a value of x if one or more of the measures is given as an expression in terms of x. In solving the equation, it is important to subtract carefully. The solution process can then be simplified by multiplying both sides of the equation by 2.

