Question

1. The circle shown below has AB and BC as its tangents:

AB and BC are two tangents to a circle which intersect outside the circle at a point B.

If the measure of arc AC is 140°, what is the measure of angle ABC?


60°
40°
70°
50°


2 .Lines CD and DE are tangent to circle A shown below:

Lines CD and DE are tangent to circle A and intersect at point D. Arc CE measures 112 degrees. Point B lies on circle A.

If Arc CE is 112°, what is the measure of ∠CDE?

124°
136°
68°
56°

3.In circle Q shown below, the mArc UR is 199° and the mArc RS is 76°.

Circle with center Q with a tangent line touching at point U and extending to point T. A secant line through RS and extending to point T.

What is the measure of ∠STU?

57°
61.5°
137.5°
142°

Answer 1

#1) 40°

#2) 68°

#3) 57°

Explanation:

The measure of an angle formed by two tangents, a tangent and a secant or two secants is equal to half the difference between the intercepted arcs.

For #1, arc AC is 140°. This makes the other arc equal to 360-140 = 220°. The difference between the arcs is 220-140 = 80°; half of this is 80/2 = 40°.

For #2, arc CE is 112°. This makes arc CBE 360-112 = 248°. The difference between the two arcs is 248-112 = 136°. Half of this is 136/2 = 68°.

For #3, arc RS is 76° and arc UR is 199°. This makes the remaining arc, SU, 360-(76+199) = 360-(275) = 85.

The measures of the intercepted arcs are then 85 and 199; the difference between them is 199-85 = 114. Half of this is 114/2 = 57°.

Answer 2

Answers
1) 40°
2) 56°
3) 57°

Explanations
Q1
Let the center of the circle to be O.
The radius meets the tangent at 90° at the point of intersection with the circumference. So the angles at A and C is 90°.
Since OABC is a quadrilateral, the angles will add up to 360°.
The angle ABC = 360° - (90°+90°+140°)
= 40°

Q2
For this question we have to apply the angle properties of a circle.
One of the properties says, "the angle subtended by the same arc at the center is doubt the angle at the circumference".
In this case the angle at the center we are told is 112°.
The angle at B (which is the angle at the circumference) is 112/2 = 56°

Q3
If arc RU=199°,
then arc RSU = 360° - 199° = 161°
The ΔQRS is an isosceles triangle where <QRS = <QSR = (180-76)÷2 = 52°
the angle at U is 90° since it is the radius meeting the tangent at U.
Now we have the quadrilateral QRSU.
So <STU = 360° - (161°+52°+90°)
= 360° - 303°
= 57°