My Math Forum  

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

Geometry Geometry Math Forum


Thanks Tree1Thanks
  • 1 Post By mathman
Reply
 
LinkBack Thread Tools Display Modes
May 15th, 2018, 02:13 PM   #1
Senior Member
 
Joined: Jan 2017
From: US

Posts: 113
Thanks: 6

Solve for the value of sin

Could anyone point me in the right direction of solving for the value of sin 150 degrees without using a calculator? I have to come up with an exact value, not a rounded decimal.


Any help would be greatly appreciated!
Indigo28 is offline  
 
May 15th, 2018, 03:04 PM   #2
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,773
Thanks: 1427

reference angle is 30 degrees in quad II, forming a 30-60-90 triangle.

what do you recall about the side ratios for such a triangle?
Attached Images
File Type: jpg 30_60_90.jpg (10.7 KB, 2 views)
skeeter is online now  
May 16th, 2018, 03:41 AM   #3
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,261
Thanks: 894

150= 90+ 60. Now, think about an equilateral triangle. It has all thee sides the same length, s, say, and all three angle with measure 60 degrees. If you drop a perpendicular from one vertex to the opposite side, it divides the equilateral triangle into to right angles with angles 60 and 30 degrees. The hypotenuse has length s, one side has length s/2 and you can calculate the third side using the Pythagorean theorem.
Country Boy is offline  
May 16th, 2018, 06:43 AM   #4
Senior Member
 
Joined: Oct 2013
From: New York, USA

Posts: 619
Thanks: 83

I used skeeter's diagram and my memory about side ratios in a 30-60-90 triangle to calculate cos 30, which equals sin 150.
EvanJ is offline  
May 16th, 2018, 01:54 PM   #5
Global Moderator
 
Joined: May 2007

Posts: 6,626
Thanks: 622

Quote:
Originally Posted by EvanJ View Post
I used skeeter's diagram and my memory about side ratios in a 30-60-90 triangle to calculate cos 30, which equals sin 150.
sin(150)=sin(30), not cos(30). Your approach still works.
Thanks from EvanJ
mathman is offline  
May 17th, 2018, 01:00 PM   #6
Senior Member
 
Joined: Oct 2013
From: New York, USA

Posts: 619
Thanks: 83

I figured out my mistake. I thought sin x = cos (180-x) when it is sin x = cos (90-x).
EvanJ is offline  
June 3rd, 2018, 02:05 PM   #7
Newbie
 
Joined: Dec 2017
From: Spain

Posts: 18
Thanks: 1

For common angles, like 0º, 30º, 45º, 60º and 90º, there are special sine-values.
Since sin(150)=sin(30)=0.5
Remember:
sin(0)= 0.5*[sqrt(0)]
sin(30)=0.5*[sqrt(1)]=0.5
sin(45)=0.5*[sqrt(2)]
sin(60)=0.5*[sqrt(3)]
sin(90)=0.5*[sqrt(4)]=1

Look in this page:
Sines and Cosines of Common Angles -- The Brain Dump
MIKI14 is offline  
June 4th, 2018, 10:09 AM   #8
Global Moderator
 
Joined: Dec 2006

Posts: 19,865
Thanks: 1833

sin(15$^\circ\!$) = 0.25(√6 - √2)
sin(18$^\circ\!$) = 0.25(√5 - 1)
sin(54$^\circ\!$) = 0.25(√5 + 1)
sin(75$^\circ\!$) = 0.25(√6 + √2)
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
sin, solve



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Please Solve this 4 EPIC Problems.... i couldn't solve it myself gen_shao Algebra 12 November 2nd, 2014 07:11 AM





Copyright © 2018 My Math Forum. All rights reserved.