AB is a diameter of a circle of radius 10 units. C is a point on the circle such that length of arc BC = \(\frac{20π}{3}\) units. The angle bisector of angle ACB cuts the circle at D. Find the length of CD.
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Let β B O C = ΞΈ
r ΞΈ = 3 2 0 Ο β βΆ 1 0 ΞΈ = 3 2 0 Ο β βΆ ΞΈ = 3 2 Ο β = 1 2 0 β
β A O C = β A C O = β C A O = 6 0 β
β A C D = β B C D = 4 5 β
So, β O C D = 1 5 β and β C O D = 1 5 0 β
cos 2 1 5 β = 2 1 + cos 3 0 β β
cos 2 1 5 β = 2 1 + 2 3 β β β
cos 2 1 5 β = 4 2 + 3 β β
c o s 2 1 5 β = 1 6 8 + 4 3 β β
c o s 2 1 5 β = 1 6 6 + 2 1 2 β + 2 β
c o s 2 1 5 β = 1 6 ( 6 β + 2 β ) 2 β
c o s 1 5 β = 4 6 β + 2 β β
Now Using Sine Law in β³ C O D ,
sin 1 5 0 β C D β = sin 1 5 β C O β
C D = sin 1 5 β sin 3 0 β Γ 1 0 β
C D = sin 1 5 β 2 sin 1 5 β cos 1 5 β Γ 1 0 β
C D = 2 cos 1 5 β Γ 1 0
C D = 5 2 β ( 3 β + 1 )