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Remember that if a quadrilateral is inscribed in a circle, the opposite angles sum up 180 degrees
So, if you sum ROP + PQR = 180
ROP=x+16 and PQR=6x-4, so ROP+PQR=x+16+6x-4=180
x+16+6x-4 = 7x+12 = 180, can you do this?
I got 12=12=180 @Natriumhydrid
Quadrilaterals add up to 360, so couldn't you add all of the equations (angle p=x) and set them equal to 360
(x+16) + (2x+16) + (6x-4) + x =360 angle P=x
I know that the opposite angles in a inscribed quadrilateral are supplementary.
So x=83 @Katherine2016
Do we apply that x to all the equations around the circle? @Katherine2016
how do you get 12=12=180?
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)°
I think its the first one?
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...
Am I right?
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
The correct is A
Can you help me with another question? @Natriumhydrid
Thank you again!
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?
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
Omg again! I forgot to put the answer choices! 29° 8° 16° 21°
is 8 degrees :)