circlesgeometry
In the given figure, PQ and PR are tangents to the circle with centre $O$ and $S$ is a point on the circle such that $\angle{SQL}={50}^{\circ}$ and $\angle{SRM}={60}^{\circ}$. Find $\angle{QSR}.$
What I've tried,
Join $OQ$ and $OR$. Since the line joining the point of contact of the tangent to the centre of the circle is equal to $90^{\circ}$.
$\therefore$$\angle$OQL=$\angle{ORL}={90}^{\circ}$
But, now I am stuck.
Following the comment by martini: since every radius of circle is perpendicular to the corresponding tangent line, the center $O$ must be such that $OP_1\perp \ell_1$ and $OP_2\perp \ell_2$. This already determines $O$ as the point of intersection of the perpendiculars to $OP_j$ passing through $P_j$, $j=1,2$.
The solution is unique, if it exists; but it does not exist when $|OP_1|\ne |OP_2|$. (The problem is overdetermined, as Hagen von Eitzen said.)
First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.
From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which is the smaller diagonal from the kite. Now we need to find the other one.
Let $D$ be the point when both diagonals intersect, and from the propertiy of the kite we know that they are normal to each other. So we can split the diagonal $PO$ into $PO = OD + PD$. We know that the $OD$ is the height of the equilaterial triangle so we have:
$$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
Draw your two radii. $\triangle SOR$ is isoceles and $\angle SRO=30^\circ=\angle RSO$. Similarly $\angle SQO=40^\circ=\angle QSO$, so $\angle QSR=70^\circ$