angle addition postulate (geometry 3_3)
DESCRIPTION
Students learn the Angle Addition Postulate and use it to solve problems.TRANSCRIPT
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The Angle Addition PostulateThe Angle Addition Postulate
You will learn to find the measure of an angle and the bisectorof an angle.
NOTHING NEW!
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1) Draw an acute, an obtuse, or a right angle. Label the angle RST.
R
TS
The Angle Addition PostulateThe Angle Addition Postulate
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1) Draw an acute, an obtuse, or a right angle. Label the angle RST.
R
TS
2) Draw and label a point X in the interior of the angle. Then draw SX.
X
The Angle Addition PostulateThe Angle Addition Postulate
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1) Draw an acute, an obtuse, or a right angle. Label the angle RST.
R
TS
2) Draw and label a point X in the interior of the angle. Then draw SX.
X
3) For each angle, find mRSX, mXST, and RST.
The Angle Addition PostulateThe Angle Addition Postulate
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1) Draw an acute, an obtuse, or a right angle. Label the angle RST.
R
TS
2) Draw and label a point X in the interior of the angle. Then draw SX.
X
3) For each angle, find mRSX, mXST, and RST.
30°
45°
The Angle Addition PostulateThe Angle Addition Postulate
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1) Draw an acute, an obtuse, or a right angle. Label the angle RST.
R
TS
2) Draw and label a point X in the interior of the angle. Then draw SX.
X
3) For each angle, find mRSX, mXST, and RST.
30°
45°
75°
The Angle Addition PostulateThe Angle Addition Postulate
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R
TS
X
30°
45°
75°
1) How does the sum of mRSX and mXST compare to mRST ?
The Angle Addition PostulateThe Angle Addition Postulate
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R
TS
X
30°
45°
75°
= mRST = 75
mXST = 30
+ mRSX = 45
1) How does the sum of mRSX and mXST compare to mRST ?
Their sum is equal to the measure of RST .
The Angle Addition PostulateThe Angle Addition Postulate
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R
TS
X
30°
45°
75°
= mRST = 75
mXST = 30
+ mRSX = 45
1) How does the sum of mRSX and mXST compare to mRST ?
2) Make a conjecture about the relationship between the two smaller angles and the larger angle.
Their sum is equal to the measure of RST .
The Angle Addition PostulateThe Angle Addition Postulate
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R
TS
X
30°
45°
75°
= mRST = 75
mXST = 30
+ mRSX = 45
1) How does the sum of mRSX and mXST compare to mRST ?
2) Make a conjecture about the relationship between the two smaller angles and the larger angle.
Their sum is equal to the measure of RST .
The sum of the measures of the twosmaller angles is equal to the measureof the larger angle.
The Angle Addition PostulateThe Angle Addition Postulate
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Postulate 3-3
Angle Addition Postulate
For any angle PQR, if A is in the interior of PQR, thenmPQA + mAQR = mPQR.
2
1
A
R
P
Q
The Angle Addition PostulateThe Angle Addition Postulate
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Postulate 3-3
Angle Addition Postulate
For any angle PQR, if A is in the interior of PQR, thenmPQA + mAQR = mPQR.
2
1
A
R
P
Q m1 + m2 = mPQR.
The Angle Addition PostulateThe Angle Addition Postulate
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Postulate 3-3
Angle Addition Postulate
For any angle PQR, if A is in the interior of PQR, thenmPQA + mAQR = mPQR.
2
1
A
R
P
Q m1 + m2 = mPQR.
There are two equations that can be derived using Postulate 3 – 3.
The Angle Addition PostulateThe Angle Addition Postulate
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Postulate 3-3
Angle Addition Postulate
For any angle PQR, if A is in the interior of PQR, thenmPQA + mAQR = mPQR.
2
1
A
R
P
Q m1 + m2 = mPQR.
There are two equations that can be derived using Postulate 3 – 3.
m1 = mPQR – m2
m2 = mPQR – m1
The Angle Addition PostulateThe Angle Addition Postulate
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Postulate 3-3
Angle Addition Postulate
For any angle PQR, if A is in the interior of PQR, thenmPQA + mAQR = mPQR.
2
1
A
R
P
Q m1 + m2 = mPQR.
There are two equations that can be derived using Postulate 3 – 3.
m1 = mPQR – m2
m2 = mPQR – m1
These equations are true no matter where A is locatedin the interior of PQR.
The Angle Addition PostulateThe Angle Addition Postulate
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2
1Y
Z
X
W
Find m2 if mXYZ = 86 and m1 = 22.
The Angle Addition PostulateThe Angle Addition Postulate
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2
1Y
Z
X
W
Find m2 if mXYZ = 86 and m1 = 22.
m2 + m1 = mXYZ Postulate 3 – 3.
The Angle Addition PostulateThe Angle Addition Postulate
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2
1Y
Z
X
W
Find m2 if mXYZ = 86 and m1 = 22.
m2 = mXYZ – m1
m2 + m1 = mXYZ Postulate 3 – 3.
The Angle Addition PostulateThe Angle Addition Postulate
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2
1Y
Z
X
W
Find m2 if mXYZ = 86 and m1 = 22.
m2 = mXYZ – m1
m2 = 86 – 22
m2 + m1 = mXYZ Postulate 3 – 3.
The Angle Addition PostulateThe Angle Addition Postulate
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2
1Y
Z
X
W
Find m2 if mXYZ = 86 and m1 = 22.
m2 = mXYZ – m1
m2 = 86 – 22
m2 = 64
m2 + m1 = mXYZ Postulate 3 – 3.
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
2x + (5x – 6) = 120 Substitution
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
2x + (5x – 6) = 120 Substitution
7x – 6 = 120 Combine like terms
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
2x + (5x – 6) = 120 Substitution
7x – 6 = 120 Combine like terms
7x = 126 Add 6 to both sides
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
2x + (5x – 6) = 120 Substitution
7x – 6 = 120 Combine like terms
7x = 126
x = 18
Add 6 to both sides
Divide each side by 7
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
2x + (5x – 6) = 120 Substitution
7x – 6 = 120 Combine like terms
7x = 126
x = 18
Add 6 to both sides
Divide each side by 7
mABC = 2x
mABC = 2(18)
mABC = 36
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
2x + (5x – 6) = 120 Substitution
7x – 6 = 120 Combine like terms
7x = 126
x = 18
Add 6 to both sides
Divide each side by 7
mABC = 2x
mABC = 2(18)
mABC = 36
mCBD = 5x – 6
mCBD = 5(18) – 6
mCBD = 90 – 6
mCBD = 84
The Angle Addition PostulateThe Angle Addition Postulate
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2x°
(5x – 6)°
B
DC
A
Find mABC and mCBD if mABD = 120.
mABC + mCBD = mABD Postulate 3 – 3.
2x + (5x – 6) = 120 Substitution
7x – 6 = 120 Combine like terms
7x = 126
x = 18
Add 6 to both sides
Divide each side by 7
mABC = 2x
mABC = 2(18)
mABC = 36
mCBD = 5x – 6
mCBD = 5(18) – 6
mCBD = 90 – 6
mCBD = 84
36 + 84 = 120
The Angle Addition PostulateThe Angle Addition Postulate
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Just as every segment has a midpoint that bisects the segment, every anglehas a ___ that bisects the angle.
The Angle Addition PostulateThe Angle Addition Postulate
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Just as every segment has a midpoint that bisects the segment, every anglehas a ___ that bisects the angle.ray
The Angle Addition PostulateThe Angle Addition Postulate
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Just as every segment has a midpoint that bisects the segment, every anglehas a ___ that bisects the angle.ray
This ray is called an ____________ .
The Angle Addition PostulateThe Angle Addition Postulate
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Just as every segment has a midpoint that bisects the segment, every anglehas a ___ that bisects the angle.ray
This ray is called an ____________ .angle bisector
The Angle Addition PostulateThe Angle Addition Postulate
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Just as every segment has a midpoint that bisects the segment, every anglehas a ___ that bisects the angle.ray
This ray is called an ____________ .angle bisector
The Angle Addition PostulateThe Angle Addition Postulate
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Definition of
an Angle Bisector
The bisector of an angle is the ray with its endpoint at thevertex of the angle, extending into the interior of the angle.
The bisector separates the angle into two angles of equalmeasure.
2
1
A
R
P
Q
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Definition of
an Angle Bisector
The bisector of an angle is the ray with its endpoint at thevertex of the angle, extending into the interior of the angle.
The bisector separates the angle into two angles of equalmeasure.
2
1
A
R
P
Q
QA is the bisector of PQR.
The Angle Addition PostulateThe Angle Addition Postulate
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Definition of
an Angle Bisector
The bisector of an angle is the ray with its endpoint at thevertex of the angle, extending into the interior of the angle.
The bisector separates the angle into two angles of equalmeasure.
2
1
A
R
P
Q
m1 = m2
QA is the bisector of PQR.
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1 + 2 = CAN Postulate 3 - 3
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1 + 2 = CAN Postulate 3 - 3
1 + 2 = 130 Replace CAN with 130
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1 + 2 = CAN Postulate 3 - 3
1 + 2 = 130 Replace CAN with 130
1 + 1 = 130 Replace 2 with 1
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1 + 2 = CAN Postulate 3 - 3
1 + 2 = 130 Replace CAN with 130
1 + 1 = 130 Replace 2 with 1
2(1) = 130 Combine like terms
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1 + 2 = CAN Postulate 3 - 3
1 + 2 = 130 Replace CAN with 130
1 + 1 = 130 Replace 2 with 1
2(1) = 130 Combine like terms
(1) = 65 Divide each side by 21
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1 + 2 = CAN Postulate 3 - 3
1 + 2 = 130 Replace CAN with 130
1 + 1 = 130 Replace 2 with 1
2(1) = 130 Combine like terms
(1) = 65 Divide each side by 2
Since 1 = 2
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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If bisects CAN and mCAN = 130, find 1 and 2.AT
ATSince bisects CAN, 1 = 2.
1 + 2 = CAN Postulate 3 - 3
1 + 2 = 130 Replace CAN with 130
1 + 1 = 130 Replace 2 with 1
2(1) = 130 Combine like terms
(1) = 65 Divide each side by 2
Since 1 = 2, 2 = 65
1
2
AC
N
T
The Angle Addition PostulateThe Angle Addition Postulate
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The Angle Addition PostulateThe Angle Addition Postulate