Question

In a circle with centre P, AB and CD are equal chords. If
APB = 100°, then find PCD.​

Answer 1

$\angle PCD = 40$°

Step-by-step explanation:

In a circle with centre P, AB and CD are equal chords

if two chords are equal ,then they are congruent also

$\angle APB = \angle CPD = 100$° (vertical opposite angles)

now in $\bigtriangleup PCD$

PC = PD (radii of the circle)

therefore ,

$\angle PCD = \angle PDC$ = x..(angles opposite to the equal sides are equal)

$\angle PCD + \angle CPD + \angle PDC$ = 180°  ...(angle sum property)

100 +x+x = 180°

2x= 180 - 100

2x = 80°

x= 40 °

$\angle PCD = 40$°

hence,

$\angle PCD = 40$°

#Learn more:

Ab and CD are equal chords of a circle with Centre p if angle APB=80° then find angle CPD​

https://brainly.in/question/8937252

Answer 2

Answer:

In a circle with centre P, AB and CD are equal chords

if two chords are equal ,then they are congruent also

\angle APB = \angle CPD = 100∠APB=∠CPD=100 ° (vertical opposite angles)

now in \bigtriangleup PCD△PCD

PC = PD (radii of the circle)

therefore ,

\angle PCD = \angle PDC∠PCD=∠PDC = x..(angles opposite to the equal sides are equal)

\angle PCD + \angle CPD + \angle PDC∠PCD+∠CPD+∠PDC = 180° ...(angle sum property)

100 +x+x = 180°

2x= 180 - 100

2x = 80°

x= 40 °

\angle PCD = 40∠PCD=40 °

hence,

\angle PCD = 40∠PCD=40 °