In the given figure, PA and PB are tangents to the given circle such that PA = 5 and \angle APB = 60^o The length of chord AB is

(a) $5 \sqrt{2}$ cm (b) 5 cm

(a) $5 \sqrt{3}$ cm (a) 7.5 cm

Open in App

Solution

Given: PA and PB are tangents of a circle, PA = 5 cm and ∠APB = 60°

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 ΔPACΔPBC (By SAS congruency rule) ..........(1)

∴ AC = BC

Also ∠APB = ∠APC + ∠BPC

$angle A P C equals 1 half angle A P B space space space space space space space space left parenthesis angle A P C equals angle B P C right parenthesis space space space equals 1 half cross times 60 degree equals 30 degree angle A C P plus angle B C P equals 180 degree angle A C P equals 1 half cross times 180 degree space space space space space space space space space space space space space space space left parenthesis angle A C P plus angle B P C right parenthesis space space space equals 90 degree N o w space i n space r i g h t space t r i a n g l e space A C P sin space 30 degree equals fraction numerator A C over denominator A P end fraction 1 half equals fraction numerator A C over denominator 5 end fraction A C equals 5 over 2$
AB = AC + BC = AC + AC (AC = BC)

⇒ AB =

Suggest Corrections

Similar questions

5 \sqrt{2} cm

5 \sqrt{3} cm

= 10

∠

= 60^{o}

Join BYJU'S Learning Program