Quadrilateral OPQR is inscribed in circle N, as shown below. Which of the following could be used to calculate the measure of ∠OPQ?

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- yeval76

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- yeval76

##### 1 Attachment

- anonymous

Remember that if a quadrilateral is inscribed in a circle, the opposite angles sum up 180 degrees

- anonymous

So, if you sum ROP + PQR = 180

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## More answers

- anonymous

ROP=x+16 and PQR=6x-4, so ROP+PQR=x+16+6x-4=180

- anonymous

x+16+6x-4 = 7x+12 = 180, can you do this?

- yeval76

I got 12=12=180 @Natriumhydrid

- anonymous

Quadrilaterals add up to 360, so couldn't you add all of the equations (angle p=x) and set them equal to 360

- anonymous

(x+16) + (2x+16) + (6x-4) + x =360
angle P=x

- yeval76

I know that the opposite angles in a inscribed quadrilateral are supplementary.

- yeval76

So x=83 @Katherine2016

- yeval76

Do we apply that x to all the equations around the circle? @Katherine2016

- anonymous

how do you get 12=12=180?

- yeval76

Omg I didn't put the answer choices:
m∠OPQ + (2x + 16)° = 180°
m∠OPQ = (6x − 4)° + (2x + 16)°
m∠OPQ + (x + 16)° + (6x − 4)°= 360°
m∠OPQ = (x + 16)° + (6x − 4)°

- yeval76

I think its the first one?

- yeval76

Because, since the opposite angles are supplementary, they would equal 180 so angle OPQ plus its opposite angle which is (2x+16) would be 180...

- yeval76

Am I right?

- anonymous

7x+12 = 180
7x = 168
x=24, so the x in the image is 24, replacing in ∠ORQ we get ∠ORQ=64, so ∠OPQ must be supplementary then ∠OPQ=116

- anonymous

The correct is A

- yeval76

Can you help me with another question? @Natriumhydrid

- yeval76

Thank you again!

- anonymous

ya

- yeval76

The figure below shows a triangle with vertices A and B on a circle and vertex C outside it. Side AC is tangent to the circle. Side BC is a secant intersecting the circle at point X:
What is the measure of angle ACB?

##### 1 Attachment

- anonymous

OK, so, if angle ABX=42, then the arc AX=84 (because is inscribed angle), and the arc AB=100, now using the formula of external angle, angle ACB=(100-84)/2=16/2=8, so ACB=8degrees

- yeval76

Omg again! I forgot to put the answer choices!
29°
8°
16°
21°

- anonymous

is 8 degrees :)

- yeval76

Thanks! @Natriumhydrid

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