Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 1013299
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 16, 20262026-05-16T10:04:48+00:00 2026-05-16T10:04:48+00:00

Let’s say I have an (x,y) that is always the same for the start

  • 0

Let’s say I have an (x,y) that is always the same for the start point of a line and an (x,y) that changes for the end point of that same line. The line is also always 40px long. At the start of the program the line originates in a vertical orientation (lets call it 0 degrees). Based on a user input I need the line to be redrawn a specific number of degrees from its origin by changing only the end (x,y).

SOME MORE FOOD FOR THOUGHT IF YOU NEED IT:

I’m in a rut trying to calculate this and make it work in Java. I can make the math work to calculate the point based on the arc length of a circle segment, but I don’t know how to make Java do it.

I think it would work easier based off a triangle angles since I will always know the length of two sides of a triangle (one side formed by the 40px long line and the other side formed by the start point of that line and the border of the JPanel) and the angle those two lines form. Still, my brain is mush from trying to figure it out. Any help would be much appreciated.

UPDATE:

@casablanca got me on the right track. I brushed up on my trig functions and here is how I made it work.

First off, I didn’t realize that 90 degrees was straight up, but once I did realize that I made my solution reflect that fact. I was drawing my line starting at the bottom center of the frame going out. Since the opposite side of the triangle is on the right side of the screen when the angle given by my user is less than 90 degrees and is on the left side of the screen when the angle given by my user is greater than 90 degrees I had to adjust the formula to account for that fact, thus I have four methods, one for the x coordinate on the left side of the screen (when the user given angle is greater than 90 degrees), one for the y coordinate on the left side of the screen (when the user given angle is greater than 90 degrees) and the same thing for the right side of the screen when the user given angle is less than 90 degrees. The int length in all methods is the length of the hypotenuse. Thanks again for your help @casablanca!

public double leftSideX(double angle, int length){
    double x = frameWidth/2 - (length * Math.cos(Math.toRadians(90-(Math.toDegrees(angle)-90))));
    return x;
}

public double leftSideY(double angle, int length){
    double y = frameHeight - (length * Math.sin(Math.toRadians(90-(Math.toDegrees(angle)-90))));
    return y;
}

public double rightSideX(double angle, int length){
    double x = frameWidth/2 + (length * Math.cos(angle));
    return x;
}

public double rightSideY(double angle, int length){
    double y = frameHeight - (length * Math.sin(angle));
    return y;
}
  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-16T10:04:48+00:00Added an answer on May 16, 2026 at 10:04 am

    Is this what you’re looking for?

    startX = x;
    startY = y;
    endX   = x + 40 * Math.sin(angle);
    endY   = y + 40 * Math.cos(angle);
    

    And draw a line from (startX, startY) to (endX, endY) in whatever API you’re using.

    Also note that angle is in radians. If you had it in degrees, you need to convert it first:

    angle = angle * Math.PI / 180;
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

Let's say that I have an arbitrary string like `A man + a plan
Let's say I have a table that looks something like this: ------------------------------- id|column2|column3 |column4
Let's say that I'm currently designing an application where I will need to use
Let's say I have a link in a table like: <td class=ms-vb width=100%> <a
Let's say I have a class like this: public class Person { private String
Let's say I have this: public DefaultListModel model = new DefaultListModel(); how do i
Let's say I have two files.. I want to compare them side-by-side and see
Let's say I have a row: 一天吃一個蘋果 Someone enters as a query: 天蘋 Should
Let's say I have two entities: Physician Credentials And a physician can have many
Let presume we have something like this: <div1> <h1>text1</h1> <h1>text2</h1> </div1> <div2> <h1>text3</h1> </div2>

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.