overleftrightarrow{PA} and overleftrightarrow | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In the circle $\odot(O, r)$,$PA$ and $PB$ are tangents drawn from an external point $P$ to the points of contact $A$ and $B$.By the properties of

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) In the circle $\odot(O, r)$,$PA$ and $PB$ are tangents drawn from an external point $P$ to the points of contact $A$ and $B$.By the properties of tangents,the radius is perpendicular to the tangent at the point of contact. Therefore,$\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.In the quadrilateral $OAPB$,the sum of the interior angles is $360^\circ$.Thus,$\angle AOB + \angle OAP + \angle OBP + \angle APB = 360^\circ$.Substituting the given values: $100^\circ + 90^\circ + 90^\circ + \angle APB = 360^\circ$.$280^\circ + \angle APB = 360^\circ$,which gives $\angle APB = 80^\circ$.The line $OP$ bisects $\angle APB$. Therefore,$m\angle OPB = \frac{1}{2} \angle APB = \frac{1}{2} 	imes 80^\circ = 40^\circ$.

$\overleftrightarrow{PA}$ and $\overleftrightarrow{PB}$ are tangents to the circle $\odot(O, r)$ at $A$ and $B$ respectively. If $m\angle AOB = 100^\circ$,then $m\angle OPB = \dots$ (in $^\circ$)

Similar Questions

$\overline{AB}$ is a diameter of $\odot(O, 15)$. $A$ tangent is drawn from $B$ to $\odot(O, 9)$ which touches $\odot(O, 9)$ at $D$. $\overrightarrow{BD}$ intersects $\odot(O, 15)$ at $C$. Find $AC$.

In $\triangle ABC$,$m \angle B = 90^\circ$,$AB = 4$ and $BC = 3$. Find the radius of the incircle of the triangle.

Radii of two concentric circles are $13$ and $8$. $\overline{AB}$ is a diameter of the circle with the larger radius. $\overline{BD}$ touches the circle with the smaller radius at $D$. Find $AD$.

If the measure of the angle between two radii of the circle is $48^{\circ}$,then the measure of the angle between the tangents drawn at the end points of the radii is $\ldots \ldots \ldots$ (in $^{\circ}$)

If from an external point $B$ of a circle with centre $O$,two tangents $BC$ and $BD$ are drawn such that $\angle DBC = 120^{\circ}$,prove that $BC + BD = BO$,i.e.,$BO = 2BC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module