Question

in ∆ABC , angle CAB is an obtuse angle . P is circumcentre of ∆ABC . prove that angle PBC = ANGLE CAB -90°

Answer 1

Answer:

∠BPC = ∠CAB - 90°

Step-by-step explanation:

Given as :

In Triangle ABC ,

∠CAB is obtuse angle

P is circumcentre Of ΔABC .

To prove that ∠PBC = ∠CAB - 90°

∠BPC at the centre and ∠BAC at point A of the remaining circumference

∴ ∠BPC = 2 ∠BAC

∵ ∠CAB is obtuse angle

As  ∠BAC is the angle of major segment , so it is acute .

So, ∠BAC is less than 90°          (acute angle)

∴ ∠BPC = ∠CAB - 90°

Hence, ∠BPC = ∠CAB - 90° proved