mm2g3b understand and use properties of central, inscribed, and related angles

Post on 02-Jan-2016

40 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

DESCRIPTION

Standard. MM2G3b Understand and use properties of central, inscribed, and related angles. Theorem 6.9. The measure of an inscribed angle is one half the measure of its intercepted arc. Find the indicated measure in P. a. m T. b. mQR. a. M T = mRS = (48 o ) = 24 o. - PowerPoint PPT Presentation

TRANSCRIPT

MM2G3bMM2G3bUnderstand and use Understand and use

properties of central, properties of central, inscribed, and inscribed, and related anglesrelated angles

Theorem 6.9Theorem 6.9

The measure of an The measure of an inscribed angle is one inscribed angle is one half the measure of its half the measure of its

intercepted arcintercepted arc

EXAMPLE 1

a. m T mQRb.

Find the indicated measure in P.

SOLUTION

12

12

M T = mRS = (48o) = 24oa.

mQR = 180o mTQ = 180o 100o = 80o. So, mQR = 80o. – –

mTQ = 2m R = 2 50o = 100o. Because TQR is a semicircle,b.

Theorem 6.10

If two inscribed angles of If two inscribed angles of a circle intercept the a circle intercept the same arc, then the same arc, then the

angles are congruentangles are congruent

EXAMPLE 2

Find mRS and m STR. What do you notice about STR and RUS?

SOLUTION

From Theorem 6.9, you know that mRS = 2m RUS = 2 (31o) = 62o.

Also, m STR = mRS = (62o) = 31o. So, STR RUS.12

12

EXAMPLE 3

SOLUTION

Notice that JKM and JLM intercept the same arc, and so JKM JLM by Theorem 6.10. Also, KJLand KML intercept the same arc, so they must also be congruent. Only choice C contains both pairs of angles.

GUIDED PRACTICE

Find the measure of the red arc or angle.

1.

SOLUTION

m G = mHF = (90o) = 45o12

12

a.

GUIDED PRACTICE

Find the measure of the red arc or angle.

2.

SOLUTION

mTV = 2m U = 2 38o = 76o. b.

GUIDED PRACTICE

Find the measure of the red arc or angle.

3.

SOLUTION

ZYN ZXN

ZXN 72°

Notice that ZYN and ZXN intercept the same arc, and so ZYN by Theorem 6.10. Also, KJL and KML intercept the same arc, so they must also be congruent.

ZXN

Theorem 6.11Theorem 6.11

If a right triangle is If a right triangle is inscribed in a circle (all inscribed in a circle (all

vertices lie on the circle) vertices lie on the circle) then the hypotenuse is then the hypotenuse is

the diameter of the the diameter of the circle.circle.

EXAMPLE 4

PhotographyYour camera has a 90o field of vision and you want to photograph the front of a statue. You move to a spot where the statue is the only thing captured in your picture, as shown. You want to change your position. Where else can you stand so that the statue is perfectly framed in this way?

EXAMPLE 4

SOLUTION

From Theorem 6.11, you know that if a right triangle is inscribed in a circle, then the hypotenuse of the triangle is a diameter of the circle. So, draw the circle that has the front of the statue as a diameter. The statue fits perfectly within your camera’s 90o field of vision from any point on the semicircle in front of the statue.

Theorem 6.12Theorem 6.12

A quadrilateral can be A quadrilateral can be inscribed in a circle if inscribed in a circle if

and only if its opposite and only if its opposite angles are angles are

supplementary.supplementary.

EXAMPLE 5

Find the value of each variable.

a.

SOLUTION

PQRS is inscribed in a circle, so opposite angles are supplementary.

a.

m P + m R = 180o

75o + yo = 180o

y = 105

m Q + m S = 180o

80o + xo = 180o

x = 100

EXAMPLE 5

b. JKLM is inscribed in a circle, so opposite angles are supplementary.

m J + m L = 180o

2ao + 2ao = 180o

a = 45

m K + m M = 180o

4bo + 2bo = 180o

b = 30

4a = 180 6b = 180

Find the value of each variable.

b.

SOLUTION

GUIDED PRACTICE

4.

Find the value of each variable.

SOLUTION

y = 112 x = 98

GUIDED PRACTICE

5.

Find the value of each variable.

SOLUTION

c = 62 x = 10

top related