Results 1 to 4 of 4

Math Help - Weird Trig problem about exact values of sin, cos, and tan

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    4

    Weird Trig problem about exact values of sin, cos, and tan

    In this assignment of mine, I'm given an isosceles triangle with sides A, B, and C and <ACB = 90 degrees.

    It's an isoceles triangle, so the other two angles are 45 degrees each. But then the problem says "Find the exact values of sin A, cos A, and tan A, based on geometric principles; do not use a calculator."

    How do I go about answering this? According to my calculator, sin(45) and cos(45) = .707, while tan(45) = 1--great, so now what? How would I go about figuring that out without a calculator...?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member WhoCares357's Avatar
    Joined
    Feb 2009
    From
    SC
    Posts
    88
    Quote Originally Posted by paradise737 View Post
    In this assignment of mine, I'm given an isosceles triangle with sides A, B, and C and <ACB = 90 degrees.

    It's an isoceles triangle, so the other two angles are 45 degrees each. But then the problem says "Find the exact values of sin A, cos A, and tan A, based on geometric principles; do not use a calculator."

    How do I go about answering this? According to my calculator, sin(45) and cos(45) = .707, while tan(45) = 1--great, so now what? How would I go about figuring that out without a calculator...?
    http://id.mind.net/~zona/mmts/miscel...90/t454590.gif

    Those are the ratios. Use sin(45) to say (Side adjacent)/(Hypotenuse).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    4
    Quote Originally Posted by WhoCares357 View Post
    http://id.mind.net/~zona/mmts/miscel...90/t454590.gif

    Those are the ratios. Use sin(45) to say (Side adjacent)/(Hypotenuse).
    Thanks, that really helps--I'm not sure why I didn't make the connection that .707 was the square root of 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by paradise737 View Post
    Thanks, that really helps--I'm not sure why I didn't make the connection that .707 was the square root of 1.
    I believe that what you intended to type was "0.707 is HALF the square root of 2"

     .707 = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}

    .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trig exact values
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 24th 2011, 12:50 AM
  2. Exact values using trig identities
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: April 29th 2009, 05:27 PM
  3. Trig Exact Values
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 3rd 2008, 03:04 PM
  4. trig functions and exact values
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: July 2nd 2008, 05:58 AM
  5. Trig exact values
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 23rd 2008, 02:40 AM

Search Tags


/mathhelpforum @mathhelpforum