Question

In triangle abc,angle a is an obtuse angle. p is the circumcenter of triangle abc prove that

Answer 1

Given question is incomplete and the complete question is,

In triangle ABC, angle A is an obtuse angle.P is circumcentre of triangle ABC. Then prove that -anglePBC=angle A-90degree.

I can solve this question.

• Construction:
• Join PA, PB and PC.
• Given,
• ∠A is an obtuse angle.
• P is the circumcenter of ΔABC
• Proof:
• ⇒ PA = PB = PC (radii)
• In ΔPAB,
• PA=PB (radii)
• ∠PAB = ∠PBA  (angles opposite to equal sides are equal)
• In ΔPBC,
• PB=PC (radii)
• ∠PBC = ∠PCB  (angles opposite to equal sides are equal)
• ∴∠PAB = ∠PBC + ∠ABC ............(i)
• In ΔPAC,
• PA=PC (radii)
• ∠PAC = ∠PCA  (angles opposite to equal sides are equal)
• ∴∠PAC = ∠PCB +∠ACB..............(ii)
• upon adding (i) and (ii), we get,
• ∠PAB+∠PAC = ∠PBC + ∠ABC+ ∠PCB+∠ACB
• ∠BAC = ∠PBC + ∠ABC+ ∠PBC+∠ACB   (as ∠PBC = ∠PCB)
• ∠BAC = 2∠PBC + ∠ABC+ ∠ACB
• ∠BAC = 2∠PBC + 180° - ∠BAC
• 2∠BAC=2∠PBC +180°

• ⇒∠PBC = ∠BAC - 90°
• ∴∠PBC = ∠A - 90°
• Hence the proof.

Answer 2

Proved that ∠PBC =  ∠A - 90°

Given :

In triangle ABC, ∠A is an obtuse angle. P is the circum-center of triangle.

To prove : ∠PBC = ∠90°

From the figure,

Note : Figure is attached below

In ΔABC, ∠A is an obtuse angle.

Let P be the circum-center of ΔABC.

PA = PB = PC [ ∵ radi of the same circle ]

In ΔPBC,

PB = PC    [∵ radi of the same circle ]

∠PBC = ∠PCB  [ ∵ opposite angles are equal ]

In ΔPAB,

PA = PB [ ∵ radi of the same circle ]

∠PAB = ∠PBA   [ ∵ opposite angles are equal ]

∠PAB = ∠PBC + ∠ABC   -----> ( 1 )

In ΔPAC,

PA = PC   [ ∵ radi of the same circle ]

∠PAC = ∠PCA   [ ∵ opposite angles are equal ]

∠PAC = ∠PCB + ∠ACB   -----> ( 2 )

Add equation ( 1 ) and ( 2 ), we get

∠PAB + ∠PAC  = ∠PBC + ∠ABC  + ∠PCB + ∠ACB

∠BAC = ∠PBC + ( ∠ABC + ∠ACB ) + ∠PBC   [ ∵ ∠PBC = ∠PCB ]

∠BAC = 2.∠PBC + ( 180° - ∠BAC )

∠BAC = 2.∠PBC + 180° - ∠BAC

∠BAC +  ∠BAC =  2.∠PBC + 180°

2. ∠BAC  = 2.∠PBC + 180°

2. ∠BAC  = 2 (∠PBC + 90°)

Number "2" on left-hand-side and right-hand-side get cancel each other.

∠BAC  = ∠PBC + 90°

∠PBC =  ∠BAC - 90°

∠PBC =  ∠A - 90°

Hence, proved that ∠PBC =  ∠A - 90° .

To learn more...

1. In triangle ABC, angle A is an obtuse angle.p is the circumcentre of triangles ABC of triangles ABC . Provethat angle PBC = angleA-90°

brainly.in/question/14988251

2. Angle CAB of Triangle ABC is obtuse angle. P is circumcentre  of Triangle ABC.prove that angle PBC=angleCAB-90

brainly.in/question/14988261