what is the angle subtended at the centre of a circle by radius 5cm by a cord of length 5cm ?
Answers
Step-by-step explanation:
We are given that a circle of radius 5 cm and chord of length 5 cm
We have to find the angle subtended at the centre of circle
Let O be the center of given circle and \thetaθ be the angle subtended at the centre by a chord
OA=OB=5 cm
AB= 5cm
Draw a OP is perpendicular bisector to AB
Then AP=PB and angle AOP=BOP=\frac{\theta}{2}
2
θ
AP=PB=\frac{5}{2} cm
2
5
cm
We know that
Sin\theta=\frac{Perpendicular }{Hypotenuse}Sinθ=
Hypotenuse
Perpendicular
Sin \frac{\theta}{2}=\frac{AP}{AO}Sin
2
θ
=
AO
AP
sin \frac{\theta}{2}=\frac{\frac{5}{2}}{5}sin
2
θ
=
5
2
5
sin\frac{\theta}{2}=\frac{5}{2\times 5}sin
2
θ
=
2×5
5
sin \frac{\theta}{2}=\frac{1}{2}sin
2
θ
=
2
1
We know that sin 30^{\circ}=\frac{1}{2}sin30
∘
=
2
1
Therefore,sin\frac{\theta}{2}=sin 30^{\circ}sin
2
θ
=sin30
∘
\frac{\theta}{2}=30^{\circ}
2
θ
=30
∘
\theta=30\times 2=60^{\circ}θ=30×2=60
∘
Hence, the angle subtended at the center by the chord =60^{\circ}60
∘