My Math Forum How ratios are written here?
 User Name Remember Me? Password

 Geometry Geometry Math Forum

 April 9th, 2018, 09:33 AM #1 Senior Member   Joined: Aug 2014 From: India Posts: 343 Thanks: 1 How ratios are written here? In the below figure, O is the center of the circle. Radius is 7 cm. $\angle$ BOC = 120 Find the length of BC? Solution: Join AO to meet BC in M $\triangle ABC \cong \triangle AOC$ $\triangle BMO \cong \triangle CMO$ $\angle BMO$ = 90 Triangle BMO has three angles are 90, 60 and 30. The sides are in the ratio of 2: $\sqrt 3$ : 1 2X = 7 $\Rightarrow$ X = 7/2 BM = $\Rightarrow$ $\sqrt 3$ x 7 / 2 = 7$\sqrt 3 / 2$ BC = 2 x 7$\sqrt 3 / 2$ = 7 $\sqrt 3$ My Question is: How ratios are written here? Last edited by Ganesh Ujwal; April 9th, 2018 at 09:36 AM.
 April 9th, 2018, 01:20 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449 Latex input \$\dfrac{a}{b}\$ will yield $\dfrac{a}{b}$
 April 9th, 2018, 03:29 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 Kite.PNG Triangle BMO is half of an equilateral triangle, so MO = 3.5cm. Now apply Pythagoras to find BM.
April 9th, 2018, 07:36 PM   #4
Senior Member

Joined: Aug 2014
From: India

Posts: 343
Thanks: 1

Quote:
 Originally Posted by skeeter Latex input \$\dfrac{a}{b}\$ will yield $\dfrac{a}{b}$
Are you trolling ? I know how ratios are written in tex command.

April 9th, 2018, 07:43 PM   #5
Senior Member

Joined: Aug 2014
From: India

Posts: 343
Thanks: 1

Quote:
 Originally Posted by skipjack Attachment 9579 Now apply Pythagoras to find BM.
Why you think $\angle$ BMO is right angled triangle?

Don't tell me I told $\angle$ BMO is 90. Use your own method and prove me.

April 10th, 2018, 12:27 AM   #6
Math Team

Joined: May 2013
From: The Astral plane

Posts: 1,979
Thanks: 789

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by Ganesh Ujwal Are you trolling ? I know how ratios are written in tex command.
Then what are you asking??

-Dan

 April 10th, 2018, 02:33 AM #7 Senior Member   Joined: Aug 2014 From: India Posts: 343 Thanks: 1 How is the ratio obtained at the step shown in the solution?
 April 10th, 2018, 04:04 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,098 Thanks: 1905 The diagram is symmetrical about AO, so angles BMO and CMO have equal measure, which implies they are right angles; also angles BOM and COM have equal measure, which implies they are both 60$^\circ$. Knowing the angles of triangle BMO are 90$^\circ$, 60$^\circ$ and 30$^\circ$ implies that it is half of an equilateral triangle.
April 10th, 2018, 08:33 AM   #9
Senior Member

Joined: Aug 2014
From: India

Posts: 343
Thanks: 1

Quote:
 Originally Posted by skipjack The diagram ....... triangle.
My question is different. How ratios are obtained?

I got the answer myself: http://mathcentral.uregina.ca/qq/dat....05/gary1.html

Can you tell me how 2X = 7 is obtained in my solution?

Last edited by Ganesh Ujwal; April 10th, 2018 at 08:46 AM.

 April 10th, 2018, 10:17 AM #10 Senior Member   Joined: Aug 2014 From: India Posts: 343 Thanks: 1 I got the answer myself: 2x = 7 because 2 x is hypotenuse. Here hypotenuse is radius of the circle i.e 7.

 Tags ratios, written

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Ganesh Ujwal Geometry 6 April 9th, 2018 09:08 AM yuceelly Trigonometry 3 December 19th, 2015 11:55 AM EvanJ Algebra 5 March 4th, 2015 10:42 AM mathman1990 Abstract Algebra 0 October 6th, 2011 05:48 AM mathmajorintrouble Abstract Algebra 11 May 4th, 2009 12:41 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.