Lesson 10-4

Example 1  Measures of Inscribed Angles
In circle P, m = 120, m = 45, and m = 105.	
Find the measures of the numbered angles.
First determine m.

m + m + m + m	= 360	Arc Addition Postulate
	120 + 45 + 105 + m	= 360	m = 120, m = 45, m = 105
	270 + m	= 360	Simplify.
	m	= 90	Subtract 270 from each side.
So, m = 90.
m1	= m
	= (45) or 22.5	m2	= m
	= (120) or 60	
m3	= m
	=  (90) or 45	m4	= m
	=  (45) or 22.5	
5 is the third angle of a triangle for which the measures of the other two angles are known.
	m5 + m1 + m3	= 180	Angle Sum Theorem
	m5 + 22.5 + 45	= 180	m1 = 22.5, m3 = 45
	m5 + 67.5	= 180	Simplify.
	m5	= 112.5	Subtract 67.5 from each side.	

Example 2  Proof with Inscribed Angles
Given:	Circle A with  and  equidistant from the center	
  , and   
Prove:	

Proof:
Statements	Reasons	
1.	Circle A with  and  equidistant from the center	1.	Given	
2.	  	2.	If two chords are equidistant from the center, then they are .	
3.	  	3.	Given	
4.	RTI  YXZ	4.	Inscribed s of  arcs are .	
5.	  	5.	Given	
6.	RIT  YZX	6.	Inscribed s of  arcs are .	
7.	rRTI  YXZ	7.	ASA	
8.		8.	CPCTC	
Example 3  Inscribed Arcs and Probability
PROBABILITY  Points M and P are on a circle so that m = 100. Suppose point N is randomly located on the same circle so that it does not coincide with M or P. What is the probability that mMNP = 50?
Since the angle measure is half the arc measure, 	
inscribed MNP must intercept , so N must not
lie on minor arc MP. Draw a figure and label any 
information you know.

m	= 360 - m
	= 360 - 100 or 260

Since MNP must intercept , the probability that mMNP = 50 is the same as the probability of N being contained in .

The probability that N is located on  is . So, the probability that mMNP = 50 is also .

Example 4  Angles of an Inscribed Triangle
ALGEBRA  Triangles EFH and FGH are inscribed	
in circle T with   . Find the measure
of each numbered angle if m2 = 3a + 2 
and m3 = 12a - 2.
FGH is a right angle because it intercepts a semicircle.
	m2 + m3 + mFGH	= 180	Angle Sum Theorem
	(3a + 2) + (12a - 2) + 90	= 180	m2 = 3a + 2, m3 = 12a - 2, mFGH = 90
	15a + 90	= 180	Simplify.
	15a	= 90	Subtract 90 from each side.
	a	= 6	Divide each side by 15.

Use the value of a to find the measures of 2 and 3.
m2	= 3a + 2	Given
	= 3(6) + 2	a = 6
	= 20	Simplify.	m3	= 12a - 2	Given
	= 12(6) - 2	a = 6
	= 70	Simplify.	

HEF is a right angle because it intercepts a semicircle.
Because   ,   , which leads to 1  4. Thus, m1 = m4.
	m1 + m4 + mHEF	= 180	Angle Sum Theorem
	m1 + m1 + 90	= 180	m1 = m4, mHEF = 90
	2(m1) + 90	= 180	Simplify.
	2(m1)	= 90	Subtract 90 from each side.
	m1	= 45	Divide each side by 2.

Since m1 = 45, m4 = 45.

Example 5  Angles of an Inscribed Quadrilateral
Quadrilateral WXYZ is inscribed in circle B. If mW  = 74 and mZ = 112, find mY and mX.
Draw a sketch of this situation.	
To find mY, we need to know m.
To find m, first find m.
m	= 2(mW)	Inscribed Angle Theorem
	= 2(74) or 148 	mW  = 74

	m + m	= 360	Sum of angles in circle = 360
	148 + m	= 360	m = 148
	m	= 212	Subtract 148 from each side.

	m	= 2(mY)	Inscribed Angle Theorem
	212	= 2(mY)	Substitution
	106	= mY	Divide each side by 2.

To find mX, we need to know m.
First, find m.
	2(mZ) = m	Inscribed Angle Theorem
	2(112) = m	mZ = 112
	224 = m	Simplify.

	m + m	= 360	Sum of angles in circle = 360
	224 + m	= 360	m = 224
	m	= 136	Subtract 224 from each side.

	m	= 2(mX)	Inscribed Angle Theorem
	136	= 2(mX)	Substitution
	68	= mX	Divide each side by 2.