Question

What is m 11 years ago

Answer 1

https://www.dekalbonlineacademy.org/content/enforced/81099-E201459727.09720011/AGM3Q3Q9.jpg?_&d2lSessionVal=KguES3n6FhYGVOPo8JyT6wZZ7

Answer 2

@GodlyMaths

Answer 3

@jim_thompson5910

Answer 4

There's an interesting type of angle on the circle, called a "inscribed angle" composed of two chords and is half of the measure of the arc it intersects:

\[<m=\frac{ arc.BC }{ 2 }\]
So, in the problem, you have an angle "m" to calculate, so by the definition and equation I gave you, m is calculated:
\[<R=\frac{ arc.QR }{ 2 }\]
arc qr is 42, so, replacing:
\[<R=\frac{ 42 }{ 2 }\]
So, now all you have to do is perform the division to obtain the answer.

Answer 5

The answer is 21 @Owlcoffee

Answer 6

correct.

Answer 7

But that is not part of my answer choices

Answer 8

Oh sorry, we calculated <P there, so I have some good news.
we have more work to do, yay!

Answer 9

Yay

Answer 10

what other work do we have to do ? @Owlcoffee

Answer 11

@Owlcoffee

Answer 12

Well, we want "R".
Apply the same property to "Q", and see what you get, the arc is 180º

Answer 13

90

Answer 14

@Owlcoffee

Answer 15

180/2 right ?

Answer 16

yes, what's the result?

Answer 17

90

Answer 18

Good, now, there's a very important property of the triangle, wich is that the sum of all the angles must be, 180º.
and the angles are <R, <Q, <P.
We have angle P and angle Q, so all we do is replace:
\[<P+<Q+<R=180º\]
\[21º+90º+<R=180º\]
Sol, try solving for angle "R"

Answer 19

How do i solve for angle R?

Answer 20

@Owlcoffee

Answer 21

@Owlcoffee

Answer 22

please help

Answer 23

just like a normal univariable equation:
\[21+90+<R=180\]
we sustract both "21" and "90" on each side:
\[\normalsize\color{red}{ -21 }\normalsize\color{red}{ -90 }+21+90+<R=180\normalsize\color{red}{ -21 }\normalsize\color{red}{ -90 }\]
and operating:
\[<R=180-21-90\]
all you have to do is do the sustraction on the right side.

Answer 24

69 is the answer

Answer 25

Thank you so much

Answer 26

at your service ;)

Answer 27

Can you help me with one more please ? I have to graduate

Answer 28

yes sure.
I'd be pleased to help you with that.