Question

Prove that chord equal to the radius of a circle subtends an angle

of 60
° at the centre of the circle.
​

Answer 1

Answer:

First method:-

Let r be the radius of a circle in which a cord AB=5cm subtends an angle of

60°at the center O. An another cord PQ which subtends an angle of 120°at

the center of the same circle.

In triangle OAB by cosine formula

cos 60°= (OA^2+OB^2-AB^2)/2.OA.OB

Putting OA=OB=r. and. AB=5 cm.

1/2=(r^2+r^2-5^2)/2.r.r

1=(2.r^2-25)/r^2=> 2r^2–25=r^2

r^2=25. =>. r = 5cm.

In triangle OPQ by cosine formula

cos120°=(OP^2+OQ^2-PQ^2)/2.OP.OQ

-1/2=(r^2+r^2-PQ^2)/2.r.r

-1= (2r^2-PQ^2)/r^2. On putting r=5 cm.

-1=(50-PQ^2)/25

-25=50-PQ^2

PQ^2=50+25=75

PQ. =√75cms. = 5√3cms. =5×1.732 =8.66 cms. Answer.

Second method:-

Let r be the radius of a circle in which a cord. AB= 5cm subtends an angle of

60° at the center O. An another cord PQ which subtends an angle of 120° at

the center of the same circle. Draw perpendiculars OC on AB and OR on

PQ.

In right angled triangle BCO

BC/OB=sin30°

(5/2)cm/r. = 1/2. =>. r= 5 cm………….(1)

In right angled triangle QRO

QR/OQ=sin60°

(PQ/2)/r = √3/2

Putting r =5 from eqn.(1)

PQ/5=√3

PQ=5.√3 cms = 5×1.732. = 8.66 cms. Answer

Step-by-step explanation: