# Calculating vectors/intercepts

1. ## Calculating vectors/intercepts

Hi all, hoping there is a trig/geometry/intercept function that could solve this problem? Or maybe I need to graph it somehow and measure the intercept?

All I know how to do right now is draw it with a ruler, compass and protractor and measure, not so accurate!
I know there's a 90 degree angle in there so possibly it simplifies the problem (?) , but it won't always be 90 degrees as I have future problems to solve with whatever solution I can develop.

Any ideas greatly appreciated!
Thanks

(NOT DRAWN TO SCALE)

2. ## Re: Calculating vectors/intercepts

There is no built in function for solving these things. As you suggest, this is first a trig problem before it really becomes a programming problem. Here's what I see:

1) With AB and AW known, I know everything there is to know about right triangle ABW (not drawn) -- including length BW.
2) With BW, CW, and angle B known, I can solve for BC using the law of cosines: https://en.wikipedia.org/wiki/Law_of_cosines Note that you will likely need the quadratic formula (https://en.wikipedia.org/wiki/Quadratic_equation ) to solve for BC.

That's how I would probably solve this, though note that I have note yet found the intersection between AB and CW. If that is an important part of the overall problem, then further application of the law of cosines and other trig functions should allow you to solve for that point.

One important thing to note -- you have you angles written in degrees. Note that Excel's built in trig functions (like other programming languages) use angles in radians. As a mathematician, I might suggest that it is often easier to learn radians and become comfortable expressing angles in radians rather than degrees. Until you are comfortable with radians, you must recognize that there are "convert degrees to radians" and "convert radians to degrees" steps whenever you are working with trig functions.

3. ## Re: Calculating vectors/intercepts

Hi kopity and welcome to the forum,

Put you problem on a X,Y Cartesian coordinate system, with A being at (0,0).

You have two circles in your problem. Look for the equations for those circles using the standard form of circles equations and solve where they intersect. The angle at point B has nothing to do with this problem.

http://www.mathwarehouse.com/geometr...f-a-circle.php

Try solving:
(x+75)^2 + y^2 = 75^2 and
x^2 + (y+25)^2 = 140^2

Then find the distance between the correct intersection of the two circles and point (-75,0).

4. Originally Posted by MrShorty
There is no built in function for solving these things. As you suggest, this is first a trig problem before it really becomes a programming problem. Here's what I see:

1) With AB and AW known, I know everything there is to know about right triangle ABW (not drawn) -- including length BW.
2) With BW, CW, and angle B known, I can solve for BC using the law of cosines: https://en.wikipedia.org/wiki/Law_of_cosines Note that you will likely need the quadratic formula (https://en.wikipedia.org/wiki/Quadratic_equation ) to solve for BC.

That's how I would probably solve this, though note that I have note yet found the intersection between AB and CW. If that is an important part of the overall problem, then further application of the law of cosines and other trig functions should allow you to solve for that point.

One important thing to note -- you have you angles written in degrees. Note that Excel's built in trig functions (like other programming languages) use angles in radians. As a mathematician, I might suggest that it is often easier to learn radians and become comfortable expressing angles in radians rather than degrees. Until you are comfortable with radians, you must recognize that there are "convert degrees to radians" and "convert radians to degrees" steps whenever you are working with trig functions.
Thanks! Yes, I don't know why I didn't pick up that I could calculate CW length and calculate CB length that way! And yes, I've already figured out the radians thing in excel! I'm very unfamiliar with rads. I've worked out how to make the conversion but it's just a funny number to me. I will look into it thanks

5. Originally Posted by MarvinP
Hi kopity and welcome to the forum,

Put you problem on a X,Y Cartesian coordinate system, with A being at (0,0).

You have two circles in your problem. Look for the equations for those circles using the standard form of circles equations and solve where they intersect. The angle at point B has nothing to do with this problem.

http://www.mathwarehouse.com/geometr...f-a-circle.php

Try solving:
(x+75)^2 + y^2 = 75^2 and
x^2 + (y+25)^2 = 140^2

Then find the distance between the correct intersection of the two circles and point (-75,0).

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