The points A, B, and C are on a circle O. The tangent line at A and the secant BC intersect at P, B lying between C and P. If $¯ ¯¯¯¯¯¯ ¯ B C$ = 20 and $¯ ¯¯¯¯¯¯ ¯ P A$ = $10 \sqrt{3}$ ,then $¯ ¯¯¯¯¯¯ ¯ P B$ equals:

5

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10

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10 \sqrt{3}

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20

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Solution

The correct option is A $10$
$¯ B C = 20 ¯ P A = 10 \sqrt{3} {¯ A P}^{2} = ¯ P B \cdot ¯ P C L e t ¯ P B = a 300 = a (a + 20) a^{2} + 20 a - 300 = 0 a = - 30 o r 10$
a cannot be negative.
$∴ a = 10$
$¯ P B = 10$ units

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The points A, B, and C are on a circle O. The tangent line at A and the secant BC intersect at P, B lying between C and P. If ¯¯¯¯¯¯¯¯BC = 20 and ¯¯¯¯¯¯¯¯PA = 10√3 ,then ¯¯¯¯¯¯¯¯PB equals:

The correct option is A 10¯BC=20¯PA=10√3¯AP2=¯PB⋅¯PCLet¯PB=a300=a(a+20)a2+20a−300=0a=−30or10a cannot be negative.∴a=10¯PB=10 units

The points A, B, and C are on a circle O. The tangent line at A and the secant BC intersect at P, B lying between C and P. If $¯ ¯¯¯¯¯¯ ¯ B C$ = 20 and $¯ ¯¯¯¯¯¯ ¯ P A$ = $10 \sqrt{3}$ ,then $¯ ¯¯¯¯¯¯ ¯ P B$ equals:

The correct option is A $10$
$¯ B C = 20 ¯ P A = 10 \sqrt{3} {¯ A P}^{2} = ¯ P B \cdot ¯ P C L e t ¯ P B = a 300 = a (a + 20) a^{2} + 20 a - 300 = 0 a = - 30 o r 10$
a cannot be negative.
$∴ a = 10$
$¯ P B = 10$ units