Suppose Q is a point on the circle with centr | vedclass

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

Question

(B) Given that $PQ = 1$ (radius of the circle) and $QR = 1$. In $	riangle PQR$,since $PQ = QR = 1$,it is an isosceles triangle. Therefore,$\angle QPR

Vedclass · Accepted Answer

$87$

Vedclass · Answer

(B) Given that $PQ = 1$ (radius of the circle) and $QR = 1$. In $	riangle PQR$,since $PQ = QR = 1$,it is an isosceles triangle. Therefore,$\angle QPR = \angle QRP = 2^{\circ}$. The sum of angles in $	riangle PQR$ is $180^{\circ}$,so $\angle RQP = 180^{\circ} - (2^{\circ} + 2^{\circ}) = 176^{\circ}$. Now,consider $	riangle S P Q$. Since $SP = SQ = 1$ (radii of the circle),it is an isosceles triangle. Thus,$\angle SQP = \angle QSP$. In $	riangle SPQ$,$\angle SPQ = 2^{\circ}$,so $\angle SQP = \frac{180^{\circ} - 2^{\circ}}{2} = \frac{178^{\circ}}{2} = 89^{\circ}$. Finally,$\angle RQS = \angle RQP - \angle SQP = 176^{\circ} - 89^{\circ} = 87^{\circ}$.

Similar Questions

The point of contact of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2-2x+2y+1=0$ is

The equation of a circle which touches both axes and the line $3x - 4y + 8 = 0$ and whose centre lies in the third quadrant is

If the ratio of the lengths of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ is $4 : 3$,then $k = \dots$

The minimum distance and maximum distance of the point $P(2,-7)$ from the circle $x^2+y^2-14x-10y-151=0$ are respectively . . . . . . units.

If the lines $12x - 5y - 17 = 0$ and $24x - 10y + 44 = 0$ are tangents to the same circle,then the radius of the circle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module