12.3 inscribed angles
DESCRIPTION
12.3 Inscribed Angles. A. B. C. Theorem 12-9:. The measure of an inscribed angles is half the measure of its intercepted arc. m B= 1 / 2 mAC. (. B. 100 0. 90 0. C. A. 60 0. 110 0. D. Example”. Find the measure of A. (. m A= 1 / 2 mBCD. m A= 1 / 2 (90 0 +60 0 ). - PowerPoint PPT PresentationTRANSCRIPT
Theorem 12-9:
The measure of an inscribed angles is half
the measure of its intercepted
arc.
mB=1/2mAC
(
B
A
C
Example”
Find the measure of A
A
B
C
D
900
1100
600
1000
mA=1/2mBCD
(mA=1/2(900+600)mA=1/2(1500)mA=750
Example”
Find the measure of D
A
B
C
D
900
1100
600
1000
mD=1/2mABC
(mD=1/2(1000+900)mD=1/2(1900)mD=950
Corollaries #1
Two inscribed angles that intercept the same arc are congruent.
mBmCB
C
Corollaries #2
An angle inscribed in a semicircle is a right angle
mB=900
B
Corollaries #3
The opposite angles of a quadrilateral inscribed in a circle are supplementary.
mA+mC=1800
mB+mD =1800B
D
A
C
Example”
Find the measure of a and b.
a
b0 O320A is inscribed in a semi-circle, a is a right
angle
Example”
Find the measure of a and b.
a
b0 O320a=900
The sum of the angles of a triangle is 1800, the other angle is 1800-900-320=580
580
Example”
Find the measure of a and b.
a
b0 O320a=900
580=1/2b580
2 21160 =b
Theorem 12-10:
The measure of an angle formed by a
tangent and a chord is half the measure of the intercepted arc.
mC=1/2mBDC
(
B
D
C
Example:
RS and TU are diameters
of A. RB is tangent to
A at point R. Find
mBRT and mTRS.
B
R
U
S
A1260
T
mBRT
B
R
U
S
A1260
T
mBRT=1/2m RT
)
mRT=mURT-mUR
) ) )
mRT=1800-1260
)
mRT=540
)
mBRT=1/2(540)mBRT=270
mTRS
B
R
U
S
A1260
T
mBRS=mBRT+mTRS
270
900=270+mTRS
630=mTRS