My Math Forum  

Go Back   My Math Forum > High School Math Forum > Trigonometry

Trigonometry Trigonometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
January 31st, 2008, 10:06 PM   #1
Newbie
 
Joined: Jan 2008

Posts: 2
Thanks: 0

Sine Rule

Consider the triangle ABC where A=30 degrees b=10 and AB is the horizontal base of the triangle.
I have a problem trying to find the smallest value of "a" for which the triangle ABC can exist and range of values of a for two triangles to exist.
I calculated B=sin^-1 [10 sin (30)]/a
I can continue plugging in numbers to find the smallest value, but is there a better way? And I really don't know how to find value for two triangles.

Last edited by skipjack; February 23rd, 2018 at 08:32 PM.
Jenty850 is offline  
 
February 1st, 2008, 05:07 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 20,919
Thanks: 2203

Quote:
Originally Posted by Jenty850
I calculated B=sin^-1 [10 sin (30)]/a.
You should have obtained B = sin^-1[10sin(30°)/a] (or 180° - sin^-1[10sin(30°)/a]), which requires 10sin(30°)/a ≤ 1, i.e., 5 ≤ a. There are two possible triangles if 5 < a < 10 (use a sketch to see why).
skipjack is online now  
Reply

  My Math Forum > High School Math Forum > Trigonometry

Tags
rule, sine



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Sine rule jiasyuen Trigonometry 4 January 31st, 2014 06:59 AM
Sine Rule Exercises jakastam Trigonometry 6 March 4th, 2013 10:49 PM
sine rule equal to 2R, proof? gelatine1 Trigonometry 5 January 2nd, 2013 08:20 AM
how do yo solve this one using chain rule with quotient rule Peter1107 Calculus 1 September 8th, 2011 10:25 AM
should i apply the product rule or quotient rule? mt055 Calculus 3 October 29th, 2009 10:58 PM





Copyright © 2019 My Math Forum. All rights reserved.