Question

investigating inscribed angles on diameters
inscribed angle of a diameter theorem
the endpoints of a diameter lie on an inscribed angle if and only if the inscribed angle is a right angle.
example
the center of the circle is c. find the measure using the appropriate theorems and postulates.
1 find (mangle mnp)
since (overline{np}) is a diameter, (mangle mnp = 90^{circ})
your turn
the center of the circle is c. find each measure using the appropriate theorems and postulates.

1. (mangle mlp)

elaborate
2 select which statements are true or false in this figure using the theorems you learned in this lesson. there are three true statements.
statement true false
(mangle asr = 90^{circ}) (square) (square)
(mangle srq = 90^{circ}) (square) (square)
(mangle sar=\frac{1}{2}cdot mwidehat{rs}) (square) (square)
(mangle sqr=\frac{1}{2}cdot mwidehat{rs}) (square) (square)

Explanation:

Step1: Recall inscribed - angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc.

Step2: Find the measure of arc MP

The sum of the measures of the arcs of a circle is 360°. The arc MN has a central - angle measure of 70° and arc NP has a central - angle measure of 60°. So, the measure of arc MP is \(70^{\circ}+60^{\circ}=130^{\circ}\).

Step3: Calculate \(m\angle MLP\)

Since \(\angle MLP\) is an inscribed angle intercepting arc MP, by the inscribed - angle theorem \(m\angle MLP=\frac{1}{2}m\widehat{MP}\). Substituting \(m\widehat{MP} = 130^{\circ}\), we get \(m\angle MLP=\frac{1}{2}\times130^{\circ}=65^{\circ}\).

Step4: Analyze statements in question 2

For \(m\angle ASR = 90^{\circ}\):

If \(AR\) is a diameter, then by the inscribed - angle of a diameter theorem, \(m\angle ASR = 90^{\circ}\) (True).

For \(m\angle SRQ = 90^{\circ}\):

There is no information indicating that the arc or chord related to \(\angle SRQ\) makes it a right - angle. So, \(m\angle SRQ=90^{\circ}\) is False.

For \(m\angle SAR=\frac{1}{2}m\widehat{RS}\):

By the inscribed - angle theorem, the measure of an inscribed angle (\(\angle SAR\)) is half the measure of its intercepted arc (\(\widehat{RS}\)), so this statement is True.

For \(m\angle SQR=\frac{1}{2}m\widehat{RS}\):

\(\angle SQR\) does not intercept arc \(RS\). So, \(m\angle SQR=\frac{1}{2}m\widehat{RS}\) is False.

Answer:

1. \(m\angle MLP = 65^{\circ}\)

2.

• \(m\angle ASR = 90^{\circ}\): True
• \(m\angle SRQ = 90^{\circ}\): False
• \(m\angle SAR=\frac{1}{2}m\widehat{RS}\): True
• \(m\angle SQR=\frac{1}{2}m\widehat{RS}\): False