1. 10.4 Other Angle Relationships in Circles

2. Learning Target
I can use theorems about tangents, chords and secants to solve unknown measure of arcs and angles .

3. Review on inscribe angles
You know that measure of an angle inscribed in a circle is half the measure of its intercepted arc. This is true even if one side of the angle is tangent to the circle. n x
mADB = ½m
2(Angle) = arc

4. Theorem 10.12
If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one half the measure of its intercepted arc. n x
m1= ½m
2(angle) = arc
m2= ½m

5. Ex. 1: Finding Angle and Arc Measures
Line m is tangent to the circle. Find the measure of the red angle or arc.
Solution:
2(m1) =
2(x) = 150°
x = 75°
150°
2(angle) = arc

6. Ex. 1: Finding Angle and Arc Measures
Line m is tangent to the circle. Find the measure of the red angle or arc.
Solution:
2(130°) = m
m = 260°
130°
2(angle) = arc

7. Ex. 2: Finding an Angle Measure
In the diagram below,
is tangent to the circle. Find mCBD
Solution:
2(mCBD) = m
2(5x) = 9x + 20
10x = 9x +20
x = 20
 mCBD = 5(20°) = 100°
(9x + 20)°
5x° D
2(angle) = arc

8. Lines Intersecting Inside or Outside a Circle
If two lines intersect a circle, there are three (3) places where the lines can intersect.
on the circle

9. Inside the circle

10. Outside the circle

11. Lines Intersecting
You know how to find angle and arc measures when lines intersect on the circle.
You can use the following theorems to find the measures when the lines intersect
INSIDE or OUTSIDE the circle.

12. Theorem 10.13 n x
If two chords intersect in the interior of a circle, then the measure of each angle is one half the sum of the measures of the arcs intercepted by the angle and its vertical angle. f
m1 = ½ m m
m2 = ½ m m
2(angle) = ( arc1 + arc 2)

13. Theorem 10.14
If a tangent and a secant, two tangents or two secants intercept in the EXTERIOR of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs. x n f
m1 = ½ m( m )
2(angle) = ( arc1 - arc 2)

14. Theorem 10.14
If a tangent and a secant, two tangents or two secants intercept in the EXTERIOR of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs. x n f
2(angle) = ( arc1 - arc 2)
2(m2) = m( m )

15. Theorem 10.14
If a tangent and a secant, two tangents or two secants intercept in the EXTERIOR of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs. x 3 n f
2(angle) = ( arc1 - arc 2)
m3 = ½ m( m )

16. Ex. 3: Finding the Measure of an Angle Formed by Two Chords
106°
Find the value of x
Solution:
2 x° (m m
2x° = (106° + 174°)
x = 140 x°
2(angle) = ( arc1 - arc 2)
174°
Apply Theorem 10.13
Substitute values
Simplify

17. Ex. 4: Using Theorem 10.14 2(angle) = ( arc1 - arc 2)
200°
2(angle) = ( arc1 - arc 2)
Find the value of x
Solution:
2(72°) = (200° - x°)
144 = x°
- 56 = -x
56 = x x°
2(mGHF = m( m )
72°
Apply Theorem 10.14
Substitute values.
Multiply each side by 2.
Subtract 200 from both sides.
Divide by -1 to eliminate negatives.

18. Ex. 4: Using Theorem 10.14 Find the value of x Solution:
Because and make a whole circle, m =360°-92°=268° x°
92°
Find the value of x
Solution:
= ½ ( )
= ½ (176)
= 88
mGHF = ½ m( m )
Apply Theorem 10.14
Substitute values.
Subtract
Multiply
Angle x = ½ ( f – n )

19. Ex. 5: Describing the View from Mount Rainier
You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level. Find the measure of the arc that represents the part of Earth you can see.

20. Ex. 5: Describing the View from Mount Rainier
You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level. Find the measure of the arc that represents the part of Earth you can see.

21. Ex. 5: Describing the View from Mount Rainier
and are tangent to the Earth. You can solve right ∆BCA to see that mCBA  87.9°. So, mCBD  175.8°. Let m = x° using Trig Ratios

22. From the peak, you can see an arc about 4°.
175.8  ½[(360 – x) – x]
175.8  ½(360 – 2x)
175.8  180 – x
x  4.2
Apply Theorem
Simplify.
Distributive Property.
Solve for x.
From the peak, you can see an arc about 4°.

23. Reminders: Pair-Share: Work on page. 624-625 #2-35
Refer to the summary sheet “Angles Related to Circles” to identify what formula to use.