Question

100°
3. In the adjoining figure,
LPQR= 100°, where P, Q and Rare
points on a circle with centre O. Find
LOPR.
[NCERT EXERCISE]
P
R​

Answer 1

Answer:

helping....

Explanation:

Here, PR is chord

Here, PR is chordWe mark s on major arc of the circle.

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o ∠PSR=180

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o ∠PSR=180 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o ∠PSR=180 o −100

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o ∠PSR=180 o −100 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o ∠PSR=180 o −100 o

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o ∠PSR=180 o −100 o ∠PSR=80

Here, PR is chordWe mark s on major arc of the circle.∴ PQRS is a cyclic quadrilateral.So, ∠PQR+∠PSR=180 o [Sum of opposite angles of a cyclic quadrilateral is 180 o ]100+∠PSR=180 o ∠PSR=180 o −100 o ∠PSR=80 o

So, ∠POR=2∠PSR

∠POR=2×80

∠POR=2×80 o

∠POR=2×80 o =160

∠POR=2×80 o =160 o

∠POR=2×80 o =160 o

∠POR=2×80 o =160 o Now,

∠POR=2×80 o =160 o Now,In ΔOPR,

∠POR=2×80 o =160 o Now,In ΔOPR,OP=OR

∴∠OPR=∠ORP

Also in ΔOPR,

Also in ΔOPR,∠OPR+∠ORP+∠POR=180

Also in ΔOPR,∠OPR+∠ORP+∠POR=180 o

Also in ΔOPR,∠OPR+∠ORP+∠POR=180 o (Angle sum property of triangle)

Also in ΔOPR,∠OPR+∠ORP+∠POR=180 o (Angle sum property of triangle)∠OPR+∠OPR+∠POR=180

Also in ΔOPR,∠OPR+∠ORP+∠POR=180 o (Angle sum property of triangle)∠OPR+∠OPR+∠POR=180 o

Also in ΔOPR,∠OPR+∠ORP+∠POR=180 o (Angle sum property of triangle)∠OPR+∠OPR+∠POR=180 o

2∠OPR+160=180

2∠OPR+160=180 o

2∠OPR+160=180 o

2∠OPR+160=180 o 2∠OPR=180

2∠OPR+160=180 o 2∠OPR=180 o

2∠OPR+160=180 o 2∠OPR=180 o −160

2∠OPR+160=180 o 2∠OPR=180 o −160 o

2∠OPR+160=180 o 2∠OPR=180 o −160 o

2∠OPR+160=180 o 2∠OPR=180 o −160 o 2∠OPR=20

2∠OPR+160=180 o 2∠OPR=180 o −160 o 2∠OPR=20∠OPR=20/2

2∠OPR+160=180 o 2∠OPR=180 o −160 o 2∠OPR=20∠OPR=20/2∴∠OPR=10

2∠OPR+160=180 o 2∠OPR=180 o −160 o 2∠OPR=20∠OPR=20/2∴∠OPR=10 o

2∠OPR+160=180 o 2∠OPR=180 o −160 o 2∠OPR=20∠OPR=20/2∴∠OPR=10 o