Question

If PA and PB are tangents from an outside point P such that PA = 6 cm and angle APB = 60° .find the length of the chord AB.

Answer 1

$\large\red{\bf{\underline{Question:-}}}$

If PA and PB are tangents from an outside point P such that PA = 6 cm and angle APB = 60° .find the length of the chord AB.

$\large\red{\bf{\underline{Solution:-}}}$

Given:-

PA and PB are tangents of a circle, PA = 10 cm and ∠APB = 60°

To find:-

Length Of Chord AB.

Step By Step Explanation:-

Let O be the center of the given circle and C be the point of intersection of OP and AB.

In ΔPAC and ΔPBC,

》PA = PB (Tangents from an external point are equal)

》∠APC = ∠BPC (Tangents from an external point are equally inclined to the segment joining center to that point)

》PC = PC (Common)

Thus, ΔPAC is congruent to ΔPBC (By SAS congruency rule) ..........(1)

∴ AC = BC

Also ∠APB = ∠APC + ∠BPC

∠APC= ${\frac{1}{2}}$ ∠APB.

[Since,∠APC = ∠BPC]

${\frac{1}{2}}$ ×60° = 30°

∠ACP + ∠BCP = 180°. {∠ACP =∠BCP}

∠ACP = ${\frac{1}{2}}$ × 180°

Now, In right triangle ACP,

sin30° = ${\frac{AC}{AP}}$

${\frac{1}{2}}$ = ${\frac{AC}{10}}$

AC = ${\frac{1}{2}}$ × 10 = 5

∴ AB = AC + BC = AC + AC (AC = BC)

⇒ AB = 5cm + 5cm

⇒ AB = 10cm.

Thanks. :)

Answer 2

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Given:-

PA and PB are tangents of a circle, PA = 10 cm and ∠APB = 60°

To find:-

Length Of Chord AB.

Step By Step Explanation:-

Let O be the center of the given circle and C be the point of intersection of OP and AB.

In ΔPAC and ΔPBC,

- PA = PB (Tangents from an external point are equal)

- ∠APC = ∠BPC (Tangents from an external point are equally inclined to the segment joining center to that point)

- PC = PC (Common)

Thus, ΔPAC is congruent to ΔPBC (By SAS congruency rule) ..........(1)

∴ AC = BC

Also ∠APB = ∠APC + ∠BPC

∠APC= {\frac{1}{2}}

2

1

∠APB.

[Since,∠APC = ∠BPC]

{\frac{1}{2}}

2

1

×60° = 30°

∠ACP + ∠BCP = 180°. {∠ACP =∠BCP}

∠ACP = {\frac{1}{2}}

2

1

× 180°

Now, In right triangle ACP,

sin30° = {\frac{AC}{AP}}

AP

AC

{\frac{1}{2}}

2

1

= {\frac{AC}{10}}

10

AC

AC = {\frac{1}{2}}

2

1

× 10 = 5

∴ AB = AC + BC = AC + AC (AC = BC)

⇒ AB = 5cm + 5cm

⇒ AB = 10cm.