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In the given figure, PA and PB are tangents from P to a circle with centre O. If ∠AOB = 130°, then find ∠APB.
- a)40°
- b)55°
- c)50°
- d)60°
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In the given figure, PA and PB are tangents from P to a circle with ce...
In AOB, by angle sum property
Angle A + Angle B + Angle O=180°
2Angle A = 50° (Triangle is isosceles triangle)
Angle A = 25°
We know Angle APB= 2Angle A
Angle APB=2*25= 50°
Angle A + Angle B + Angle O=180°
2Angle A = 50° (Triangle is isosceles triangle)
Angle A = 25°
We know Angle APB= 2Angle A
Angle APB=2*25= 50°
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In the given figure, PA and PB are tangents from P to a circle with ce...
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In the given figure, PA and PB are tangents from P to a circle with ce...
The length of PA is 8 cm and the length of PB is 6 cm, we can use the properties of tangents to find the length of OP.
First, we know that PA and PB are tangents to the circle, which means they are perpendicular to the radius drawn from the center of the circle to the point of tangency. This means that ∠PAO and ∠PBO are right angles.
We can use the Pythagorean theorem to find the length of OP. Let's assume that OP = x cm.
In right triangle PAO, we have:
PA^2 = AO^2 + OP^2
8^2 = AO^2 + x^2
64 = AO^2 + x^2
In right triangle PBO, we have:
PB^2 = BO^2 + OP^2
6^2 = BO^2 + x^2
36 = BO^2 + x^2
Since AO and BO are radii of the same circle, they have the same length. We can denote this length as r.
Therefore, we have the system of equations:
64 = r^2 + x^2
36 = r^2 + x^2
Simplifying the system of equations, we get:
r^2 + x^2 = 64 (Equation 1)
r^2 + x^2 = 36 (Equation 2)
Subtracting Equation 2 from Equation 1, we get:
(r^2 + x^2) - (r^2 + x^2) = 64 - 36
0 = 28
This equation is not possible, which means there is no solution for r and x that satisfies the conditions. Therefore, the given figure is not possible.
First, we know that PA and PB are tangents to the circle, which means they are perpendicular to the radius drawn from the center of the circle to the point of tangency. This means that ∠PAO and ∠PBO are right angles.
We can use the Pythagorean theorem to find the length of OP. Let's assume that OP = x cm.
In right triangle PAO, we have:
PA^2 = AO^2 + OP^2
8^2 = AO^2 + x^2
64 = AO^2 + x^2
In right triangle PBO, we have:
PB^2 = BO^2 + OP^2
6^2 = BO^2 + x^2
36 = BO^2 + x^2
Since AO and BO are radii of the same circle, they have the same length. We can denote this length as r.
Therefore, we have the system of equations:
64 = r^2 + x^2
36 = r^2 + x^2
Simplifying the system of equations, we get:
r^2 + x^2 = 64 (Equation 1)
r^2 + x^2 = 36 (Equation 2)
Subtracting Equation 2 from Equation 1, we get:
(r^2 + x^2) - (r^2 + x^2) = 64 - 36
0 = 28
This equation is not possible, which means there is no solution for r and x that satisfies the conditions. Therefore, the given figure is not possible.
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In the given figure, PA and PB are tangents from P to a circle with centre O. If ∠AOB = 130°, then find ∠APB.a)40°b)55°c)50°d)60°Correct answer is option 'C'. Can you explain this answer?
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