My Math Forum How to solve this problem?

 Geometry Geometry Math Forum

April 1st, 2018, 04:03 PM   #11
Senior Member

Joined: Feb 2010

Posts: 683
Thanks: 129

Quote:
 Originally Posted by Ganesh Ujwal In Triangle ABC, the internal bisector of $\angle ABC$ and the external bisector of $\angle ACB$ meet at P.
$\displaystyle \angle BAC = 2 \cdot \angle BPC$

April 1st, 2018, 07:49 PM   #12
Newbie

Joined: Mar 2018
From: australia

Posts: 15
Thanks: 0

Quote:
 Originally Posted by Denis Messy as hell, but basically correct! Now, IF triangle ABC is isosceles: angleABC = angleACB = ? angleABP = ? angleACP = ?
I am sure you can resolve this ques when you see this pic.
Attached Images
 cat.jpg (12.5 KB, 8 views)

 April 2nd, 2018, 04:59 AM #13 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 NO need for that Helen. That'll only confuse the OP.
April 2nd, 2018, 08:41 AM   #14
Senior Member

Joined: Aug 2014
From: India

Posts: 295
Thanks: 1

Quote:
 Originally Posted by helen510 I am sure you can resolve this ques when you see this pic.

 April 3rd, 2018, 09:49 AM #15 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 Forget it, Ganesh: that will not help you... Once more: CAN YOU answer: IF your triangle ABC is isosceles: angleABC = angleACB = ? angleABP = ? angleACP = ? YES or NO ?
April 5th, 2018, 07:47 PM   #16
Senior Member

Joined: Aug 2014
From: India

Posts: 295
Thanks: 1

Quote:
 Originally Posted by Denis Forget it, Ganesh: that will not help you... Once more: CAN YOU answer: IF your triangle ABC is isosceles: angleABC = angleACB = ? angleABP = ? angleACP = ? YES or NO ?

angleABC = angleACB = 40 (since you are saying triangle is isosceles)

My question why you assume triangle is isosceles instead of obtuse or equilateral etc?

 April 5th, 2018, 08:03 PM #17 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 NO. angleABC = angleACB = 70 I'll be plain: you are a joker, else you need classroom help...
April 6th, 2018, 12:53 AM   #18
Senior Member

Joined: Aug 2014
From: India

Posts: 295
Thanks: 1

Quote:
 Originally Posted by Denis NO. angleABC = angleACB = 70 I'll be plain: you are a joker, else you need classroom help...

 April 6th, 2018, 02:10 AM #19 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,902 Thanks: 883 HOW did you get that? Show your steps...
 April 6th, 2018, 06:25 AM #20 Global Moderator   Joined: Dec 2006 Posts: 19,287 Thanks: 1681 Bisectors.PNG $\small\angle$ACD = $\small\angle$ABC + 40$^\circ$ Dividing by 2, $\small\angle$PCD = $\small\angle$PBC + 20$^\circ$ Hence $\small\angle$BPC = 20$^\circ$.

 Tags problem, solve

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post muneeb977 Linear Algebra 2 January 21st, 2017 06:36 PM tahirimanov Calculus 0 January 17th, 2015 03:19 AM luie1168e Advanced Statistics 0 February 17th, 2013 07:54 AM salgat Number Theory 2 February 18th, 2010 02:45 PM brunojo Number Theory 4 January 4th, 2008 01:56 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top