# Question b5ab2

Aug 24, 2016

$a = 6$

#### Explanation:

Construct the point $C = \left(\sqrt{3} , 0\right)$

Using the Pythagorean theorem, we have

$O A = \sqrt{O {C}^{2} + A {C}^{2}} = \sqrt{3 + 1} = 2$

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 $\triangle A O C$, then, we have

$\sin \left(\angle A O B\right) = \sin \left(\angle O A C\right) = \frac{A C}{O A} = \frac{1}{2}$

As it is clear that $0 < \angle A O B < \frac{\pi}{2}$, the only possible value for $\angle A O B$ is $\frac{\pi}{6}$.
(There are multiple ways of finding this, including using the inverse sine function or calculating it from a special right triangle, however $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$ is one of the common values worth memorizing from the unit circle)

Substituting this value in, we have $\frac{\pi}{6} = \frac{\pi}{a}$, and so $a = 6$

Aug 24, 2016

$a = 6$

#### Explanation:

Given
"The acute angle"/_AOB=pi/a=theta ("say")#
The cartesian co-ordinate of the point A is$\left(\sqrt{3} , 1\right)$

If $O A = r \to \text{Radius of the circle}$

$\text{The polar coordinate of A becomes} = \left(r , \theta\right)$

$\therefore r \cos \theta = \sqrt{3} \mathmr{and} r \sin \theta = 1$

This relation gives

$\tan \theta = \frac{1}{\sqrt{3}}$

Replacing $\theta$ by $\frac{\pi}{a}$ we get

$\implies \tan \left(\frac{\pi}{a}\right) = \tan \left(\frac{\pi}{6}\right)$

$\implies \frac{\pi}{a} = \frac{\pi}{6}$

$\therefore a = 6$

Aug 25, 2016

$a = 6$

#### 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 $\Delta O A P$ we see that cathetus $A P = 1$ (this is Y-coordinate of point $A$) and hypotenuse $O A = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{1}} = 2$ (distance from origin expressed in X- and Y-coordinates).

Therefore,
$\sin \left(\angle A O P\right) = \sin \left(\angle A O B\right) = \frac{1}{2}$
from which $\angle A O B = \frac{\pi}{6}$.

Since $\angle A O B = \frac{\pi}{a}$, we conclude that $a = 6$.