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If O is the center of the circle with diameter
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16 Jun 2021, 02:43
16 Jun 2021, 02:43
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Question Stats:
88% (03:20) correct
11% (03:22) wrong based on 18 sessions
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If O is the center of the circle with diameter AB and T is a tangency point, what is the value of x?
(A) 10
(B) 15
(C) 18
(D) 20
(E) 22.5
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Re: If O is the center of the circle with diameter
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16 Jun 2021, 05:09
16 Jun 2021, 05:09
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Carcass wrote:
If O is the center of the circle with diameter AB and T is a tangency point, what is the value of x?
(A) 10
(B) 15
(C) 18
(D) 20
(E) 22.5
Since angles in a triangle add to 180°, we know that the missing angle = (90 - 2x)°
Aside: Notice that (90 - 2x)° + 90° + 2x° =180°
Since angles on a line add to 180°, we know that the missing angle = (90 + 2x)°
Aside: Notice that (90 - 2x)° + (90° + 2x°) =180°
Next, draw a line from the center to the point of tangency. One of our circle properties tells us that this line is PERPENDICULAR to the tangent line.
Now focus on ∆ATO (below)
Since OT and OA are radii, we know that OT = OA, which makes ∆ATO an ISOSCELES triangle.
This means the other angle is ALSO (90 - 2x)°
Next, since OT is PERPENDICULAR to the tangent line, the remaining angle must be 2x°
Aside: Notice that (90 - 2x)° + 2x° =90°
Finally, we'll focus on the green triangle.
Since angles in a triangle add to 180°, we can write: x° + 2x° + (90 + 2x)° = 180°
Simplify: 5x° + 90° = 180°
Solve to get: x = 18
Answer: C
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Re: If O is the center of the circle with diameter
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16 Jun 2021, 03:55
16 Jun 2021, 03:55
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from the fig.
△ABD is right triangle
so 90+2x+ ∠TAO =180
∠TAO=90-2x
∠TAP=90+2x
draw a peripendicular from T that meets the centre of the circle
as OA=OT=radius of circle
so ∠ATO=90-2x
As ∠T =90 ∠PTA =90-(90-2x)=2x
from △PAT
x+90+2x+2x =180
5x=90
x=18
option C is correct
△ABD is right triangle
so 90+2x+ ∠TAO =180
∠TAO=90-2x
∠TAP=90+2x
draw a peripendicular from T that meets the centre of the circle
as OA=OT=radius of circle
so ∠ATO=90-2x
As ∠T =90 ∠PTA =90-(90-2x)=2x
from △PAT
x+90+2x+2x =180
5x=90
x=18
option C is correct
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