Question #b5ab2

Construct the point #C=(sqrt(3),0)#

Using the Pythagorean theorem, we have

#OA = sqrt(OC^2+AC^2)=sqrt(3+1) = 2#

The sine of an acute angle in a right triangle is equal to the length of the side opposite the angle divided by the length of the hypotenuse. Considering the right triangle #triangleAOC#, then, we have

#sin(angleAOB) = sin(angleOAC)= (AC)/(OA) = 1/2#

As it is clear that #0 < angleAOB < pi/2#, the only possible value for #angleAOB# is #pi/6#.
(There are multiple ways of finding this, including using the inverse sine function or calculating it from a special right triangle, however #sin(pi/6)=1/2# is one of the common values worth memorizing from the unit circle)

Substituting this value in, we have #pi/6 = pi/a#, and so #a=6#

Given
#"The acute angle"/_AOB=pi/a=theta ("say")#
The cartesian co-ordinate of the point A is#(sqrt3,1)#

If #OA=r->"Radius of the circle"#

#"The polar coordinate of A becomes"=(r,theta)#

#:.rcostheta=sqrt3 and rsintheta=1#

This relation gives

#tantheta =1/sqrt3#

Replacing #theta# by #pi/a# we get

#=>tan(pi/a) =tan(pi/6)#

#=>pi/a=pi/6#

#:.a=6#

Drop a perpendicular from point #A# down to X-axis. Let the base of this perpendicular on the X-axis be point #P#.

From right triangle #Delta OAP# we see that cathetus #AP=1# (this is Y-coordinate of point #A#) and hypotenuse #OA=sqrt((sqrt(3))^2+1^1)=2# (distance from origin expressed in X- and Y-coordinates).

Therefore,
#sin (/_AOP)=sin(/_AOB)=1/2#
from which #/_AOB=pi/6#.

Since #/_AOB = pi/a#, we conclude that #a=6#.