Question

In the given circle with center O two chords AB and CD are drawn. If OM and ON are perpendiculars from O to AB and CD respectively such that OM > ON and AB and CD intersected at point P, then

Answer 1

Given :  a circle with center O, two chords AB and CD are drawn which intersect at P . OM and ON ate perpendiculars from O to AB and CD respectively such that OM is greater than ON and AB and CD intersected at point P

To Find : which is true

1.angle PMN = angle PNM

2.angle PMN is greater than anglePNM

3.angle PNM is greater than angle PMN

4.angle POM=angle PON​

Solution:

OM & ON are perpendicular on AB & CD

Hence ΔPMO & ΔPNO are right angle triangles

OP = OP  ( hypotenuse common in both )

OM > ON  ( given )

PM² = OP² - OM²

PN² = OP² - ON²

=> PM² < PN²

=> PM < PN

in ΔPMN

PM < PN

=> ∠PNM <  ∠PMN

=> ∠PMN > ∠PNM

Option 2 is correct :

.angle PMN is greater than anglePNM

angle POM=angle PON​  not possible

as Cos∠POM = OM/OP  & Cos∠PON = ON/OP

OM > ON

=> Cos∠POM > Cos∠PON

=> ∠POM  < ∠PON  

Learn more:

in fig chord AB =chord CD and chord AB perpendicular chord CD .M ...

https://brainly.in/question/9085238

Find the length of the chord AC where AB and CD are the two ...

https://brainly.in/question/6652737