circlesgeometrytrigonometry
Say there are two lines that are tangent to a circle at points A and B, so that they intersect at an external point C, as shown below.
Two congruent right triangles are formed, since the tangent line is perpendicular to the radius. So line AC has the same length as line BC.
If the angle ACB is known, how can the length of line AC be calculated?
For example, if the radius $r$ is 1, and the angle ACB is $60\unicode{xb0}$, what is the length of line AC?
Note that $BCYX$ is a cyclic quadrilateral and hence $$\underbrace{\angle BCY = \angle YXA}_{\text{Exterior angle of a cyclic quadrilateral equals the interior opposite angle}}\\ \underbrace{\angle YXA = \angle TAY}_{\text{External angle of a triangle is equal to the opposite interior angle of the triangle}}$$ Hence, $$\angle BCY = \angle TAY.$$ This means the alternate angle are equal and hence the lines are parallel.
First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.
From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which is the smaller diagonal from the kite. Now we need to find the other one.
Let $D$ be the point when both diagonals intersect, and from the propertiy of the kite we know that they are normal to each other. So we can split the diagonal $PO$ into $PO = OD + PD$. We know that the $OD$ is the height of the equilaterial triangle so we have:
$$OD = \frac{r\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$
We know that the angle between the tangent line to a circle and the radius at the touching point is $90^{\circ}$, so $\angle PAO = 90^{\circ}$. We can denote $\angle PAO$ as sum of two angles: $\angle PAO = \angle OAB + \angle BAP = 60^{\circ} + \angle BAP$, because $\angle OAB$ is an angle in an equilaterial triangle. So from this we have:
$$\angle BAP = 30^{\circ}$$
Also we know that $AD = \frac r2 = \frac 12$. Now in the right triangle $PAD$ we have:
$$\tan \angle BAP = \frac{PD}{AD}$$ $$\tan \angle 30^{\circ} = \frac{PD}{\frac 12}$$ $$ \frac {1}{\sqrt{3}} = \frac{PD}{\frac 12}$$
$$ PD = \frac{1}{2\sqrt{3}}$$
Now we have: $PO = OD + PD$, i.e.:
$$PO = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}$$ $$PO = \frac{3}{2\sqrt{3}} + \frac{1}{2\sqrt{3}}$$ $$PO = \frac{4}{2\sqrt{3}}$$ $$PO = \frac{2}{\sqrt{3}} \approx 1.154$$
So the distance from $O$ to $P$ is constant, that means that $P$ can be any point that's $\frac{2}{\sqrt{3}}$ units away from $(0,0)$, i.e. $P$ lies on a circle with this equation:
$$x^2 + y^2 =\left(\frac{2}{\sqrt{3}}\right)^2$$ $$x^2 + y^2 = \frac 43$$
So the locus of the point $P$ is:
$$x^2 + y^2 = \frac 43$$
Best Answer
If $\angle ACB=60$ then $\angle ACO=30$ and therefore $OC=2$. Now you can use pythagoras and find $AC$. Using pythagoras : $OC^2=OA^2+AC^2$ and you will have: $4=1+AC^2$ Therefore $AC=\sqrt 3$