Question

In the given figure, O is the centre of the circle PS = SO = PQ/2 and PM = MO, then find thedifference between the measures of ∠QRS and ∠PQR.


Pls solve this question! and pls do not give irrelevant answers ​

Answer 1

Answer:

so it's going to be a long one, sorry i couldn't find a shorter one. But don't worry, the concept is easy.

now in triangle PSO,

PS = SO (given). ........1

OP = SO (radius) .......2

from 1 and 2

PS=OP

since all sides are equal, it's a equilateral triangle, which means all the angles the also equal (60°) .......3

PM=MO

this means, SM is median and is perpendicular to OP.

now, angle SMO =90°

angle OMR =90° ....... a

IN triangle OSM

angle OMR = 90°

angle MOS = 60° (from 3)

from these 2 angles, we can find

angle OSM(OSR)= 30°

since, OS = OR. (radius)

angle OSR = angle ORS (angles opposite to equal sides)

angle ORS = 30° ...........b

in triangle OMR, from 'a' and 'b'

angle POR = 60°

angle PQR = 30° (angle formed by a segment on any point of circle is half the angle substended by it at centre, (this therom is from class 10))

now in triangle MQR

WE, have angle QMR =90°

angle. MQR = 30°

then angle MRQ(QRS) =60°

then finally, angle QRS - angle PQR

= 60° - 30°

= 30°

and so your answer is 30°

so you may mark it as brainlist, if you are satisfied with the answer.

thank you

Answer 2

Answer:

30°

Step-by-step explanation: