Prove that chord equal to the radius of a circle subtends an angle
of 60
° at the centre of the circle.
Answers
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: