Question

in the fig.1,∆PQR=100°where P,Q and R are points on circle with centre O.find ∆oPR.

Answer 1

angle opr will be 10°
firstly take a point on the circle any where in major sector and join it with. p and r and name the point as s
now since quad. pqrs becomes cyclic therefore angle psr will be 80°
and in this we can find angle por=160°
and since por is isosceles triangle
therefore angle rpo will be 10°

Answer 2

Hello mate =_=

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Solution:

P, Q and R are the points on a circle with centre O where ∠PQR=100°

Construction: S is a point on the major arc PR. Join P and S, R and S to form a cyclic quadrilateral.

 

∠PQR+∠PSR=180°

(Sum of opposite angles of a cyclic quadrilateral is equal to 180°)

⇒100°+∠PSR=180°

⇒∠PSR=180°−100°=80°

Also, ∠POR=2∠PSR

(Angle subtended by an arc at the centre is double the angle subtended by it at the circumference of the circle.)

⇒∠POR=2×80°=160°

In ∆POR, we have

∠POR+∠ORP+∠OPR=180°

But, we have ∠ORP=∠OPR         (Angles opposite to the equal sides in a triangle are equal.)

⇒160°+∠OPR+∠OPR=180°

⇒2∠OPR=180°−160°=20°

⇒∠OPR=20/2=10°

hope, this will help you.

Thank you______❤

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