Question

Question #b5ab2

Answer 1

`a=6`

Explanation:

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`

Answer 2

`a=6`

Explanation:

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`

Answer 3

`a=6`

Explanation:

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`.