What is the length of AB if the radius of the circle with center O (shown above) is 9 and the length of the arc ACD is 3π?
A. \(9\sqrt{3}\)
B. \(9\sqrt{2}\)
C. 9
D. \(3\sqrt{2}\)
E. 3
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length of an arc = 2\(\pi\)r@/360 = 3\(\pi\)
@ = 60
which means triangle AOD = equilateral triangle
AD=OD=9
using pythagoras
\((AB)^2\) = \(9^2\) +\(9^2\) = 81(2)
AB= 9\(\sqrt{2}\)[/quote]
Apparently, Triangle OAB is assumed to be right angled (since largest side AB is considered as hypotenuse), which is not the case.
Triangle OAB is isosceles obtuse angle triangle, with angle OAB=120.
Since OA and OB and angle between them are known, one can directly compute AB by using vector algebra. Otherwise, one can prove that triangle ABD is right angles at angle DAB.
The answer is not B, but A.
Method1 (using geometry):
As calculated above @ or angle AOD=60 deg in triangle ADO
Since OD=OA=radius and angle AOD=60, hence ADO is equilateral triangle.
In triangle AOB, angle AOB= 180 - angle AOD
angle AOB=120
Also, OA=OB => angle OAB=angle OBA
By angle sum property applied at triangle AOB,
angle OAB=angle OBA=30
Now, angle DAB= angle OAD + angle OAB
=> angle DAB= 60 + 30 (triangle OAD is equilateral)
=> angle DAB=90
Hence, angle DAB is right angled triangle at angle A
Applying pythagoras theorem at triangle ADB
AB^2 + AD^2 = BD^2
AB^2= BD^2- AD^2
(BD=2*radius= 18; AD= 9)
Putting value,
AB= sqrt(243) = 9*sqrt(3)