"
">

In figure, from an external point P, two tangents PT and PS are drawn to a circle with center O and radius r. If $OP=2r$, show that $\angle  OTS =\angle OST=30^{o}$.
"

Given: From an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r and OP = 2r.

To do: To show that $\vartriangle OTS = \vartriangle OST = 30^{o}$

Solution:

In the given figure,

$OP = 2r … ( Given)$

$\angle OTP = 90^{o}$ … [radius drawn at the point of contact is perpendicular to the tangent]

In $\vartriangle OTP$,

$sin\angle OPT=\frac{OT}{OP} =\frac{1}{2}$

$=sin30^{o}$

$\Rightarrow \angle OPT=30^{o}$

$\therefore \angle TOP = 60^{o}$

$\therefore \vartriangle OTP$ is a $30^{o} -60^{o} -90^{o}$ , right triangle.

In $\vartriangle OTS$,

$OT=OS\ \dotsc \ ( Radii\ of\ the\ same\ circle)$

$\therefore \vartriangle OTS$ is an isosceles triangle.

$\therefore  \angle OTS =\angle OST \ \ \ \ \ \dotsc \ ( Angles\ opposite\ to\ equal\ sides\ of\ an\ isosceles\ triangle\ are\ equal)$

In $\vartriangle OTQ$ and $\vartriangle OSQ$

$OS=OT\ \ \ \ \ \ \dotsc \ ( Radii\ of\ the\ same\ circle)$

$OQ=OQ \ \ \ \ \ ...\ ( side\ common\ to\ both\ triangles)$

$\angle OTQ=\angle OSQ\ \ \ \ \ \dotsc \ ( angles\ opposite\ to\ equal\ sides\ of\ an\ isosceles\ triangle\ are\ equal)$

$\therefore \ vartriangle OTQ\cong \vartriangle OSQ\ \dotsc \ ( By\ S.A.S)$

$\therefore \angle TOQ= \angle SOQ=\ 60^{o}\ \dotsc \ ( C.A.C.T)$

$\therefore \angle TOS=120^{o} \dotsc \ ( \angle TOS=\angle TOQ+\angle SOQ=60^{o} +60^{o}=120^{o})$

$\therefore \angle OTS+\angle OST=180^{o}–120^{o} =60^{o}$

$\therefore  \angle OTS=\angle OST=\frac{60^{o}}{2}=30^{o}$

 

Updated on: 10-Oct-2022

235 Views

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements