x = 75 simplify

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GEOMETRY HELP x = 75 Simplify. x = mDEF Inscribed Angle Theorem 1 2 x = (mDE + mEF) Arc Addition Postulate 1 2 Because EFG is the intercepted arc of D, you need to find mFG in order to find mEFG. Find the values of x and y. x = (80 + 70) Substitute. 1 2 Inscribed Angles LESSON 12-3 Additional Examples

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1 2. x = ( 80 + 70 ) Substitute. 1 2. x = mDEF Inscribed Angle Theorem. 1 2. x = ( mDE + mEF ) Arc Addition Postulate. Because EFG is the intercepted arc of D , you need to find mFG in order to find mEFG. Inscribed Angles. LESSON 12-3. Additional Examples. - PowerPoint PPT Presentation

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Page 1: x  = 75 Simplify

GEOMETRYHELP

x = 75 Simplify.

x = mDEF Inscribed Angle Theorem 12

x = (mDE + mEF) Arc Addition Postulate12

Because EFG is the intercepted arc of D, you need to find mFG in order to find mEFG.

Find the values of x and y.

x = (80 + 70) Substitute.12

Inscribed AnglesLESSON 12-3

Additional Examples

Page 2: x  = 75 Simplify

GEOMETRYHELP

y = 95 Simplify.

y = (70 + 120) Substitute.12

y = (mEF + mFG) Arc Addition Postulate12

y = mEFG Inscribed Angle Theorem12

The arc measure of a circle is 360°, so mFG = 360 – 70 – 80 – 90 = 120.

(continued)

Quick Check

Inscribed AnglesLESSON 12-3

Additional Examples

Page 3: x  = 75 Simplify

GEOMETRYHELP

Find the values of a and b.

By Corollary 2 to the Inscribed Angle Theorem, an angle inscribed in a semicircle is a right angle, so a = 90.

Therefore, the angle whose intercepted arc has measure b must have measure 180 – 90 – 32, or 58.

Because the inscribed angle has half the measure of the intercepted arc, the intercepted arc has twice the measure of the inscribed angle, so b = 2(58) = 116.

The sum of the measures of the three angles of the triangle inscribed in O is 180. .

Inscribed AnglesLESSON 12-3

Additional Examples

Quick Check

Page 4: x  = 75 Simplify

GEOMETRYHELP

m BRT = 27 Simplify.

mRT = mURT – mUR Arc Addition Postulate

m BRT = mRT The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc (Theorem 12-10).

12

RS and TU are diameters of A. RB is tangent to A at point R. Find m BRT and m TRS.

..

m BRT = (180 – 126) Substitute 180 for m and 126 for mUR.

12

Inscribed AnglesLESSON 12-3

Additional Examples

Page 5: x  = 75 Simplify

GEOMETRYHELP

90 = 27 + m TRS Substitute.

63 = m TRS Solve.

m BRS = m BRT + m TRS Angle Addition Postulate

m BRS = 90 A tangent is perpendicular to the radius of a circle at its point of tangency.

(continued)

Use the properties of tangents to find m TRS.

Inscribed AnglesLESSON 12-3

Additional Examples

Quick Check