Trying to find Isosceles length:width ratios

[JainJude]

Hello,
I am not especially well versed in Geometry vocab, and so I'm having trouble finding the information I need.
I am an audio tech and I'd like an easy way to be more precise when placing my speakers. These speakers have a horizontal coverage expressed in degrees (typically 60-100º), that can be simplified to imagine as a circular arc. It's incredibly important that I be able to cover an area evenly while avoiding spilling into the stage area (this causes feedback, an irritating screeching/droning noise). I use a laser measure to easily mark distances on the job, and would like to be able to essentially draw myself an isosceles triangle that reflects my coverage pattern width.
Let's say that my 60º speaker is my vertex point, and one outer edge of my coverage marks the point of one base angle. From there I need to know my base length, and with that I can make accurate decisions about my coverage instead of standing behind my speaker with my hands splayed in an angle that i think looks correct.

Is there a formula that can accept a vertex angle and leg length and spit out the base length? Or, better yet, is there a clearer solution I'm not thinking of? Is doing it visually with a tool easier than any of this? I hope I've been clear.
Thanks

 

[Deleted member 4993]

Hello,
I am not especially well versed in Geometry vocab, and so I'm having trouble finding the information I need.
I am an audio tech and I'd like an easy way to be more precise when placing my speakers. These speakers have a horizontal coverage expressed in degrees (typically 60-100º), that can be simplified to imagine as a circular arc. It's incredibly important that I be able to cover an area evenly while avoiding spilling into the stage area (this causes feedback, an irritating screeching/droning noise). I use a laser measure to easily mark distances on the job, and would like to be able to essentially draw myself an isosceles triangle that reflects my coverage pattern width.
Let's say that my 60º speaker is my vertex point, and one outer edge of my coverage marks the point of one base angle. From there I need to know my base length, and with that I can make accurate decisions about my coverage instead of standing behind my speaker with my hands splayed in an angle that i think looks correct.

Is there a formula that can accept a vertex angle and leg length and spit out the base length? Or, better yet, is there a clearer solution I'm not thinking of? Is doing it visually with a tool easier than any of this? I hope I've been clear.
Thanks

Can you provide an approximate sketch of the problem - so that we can visualize the situation better?

 

[JeffM]

I am having trouble visualizing what you want to do.

Why does the triangle need to be isosceles?

In general, knowing either the length of two sides and the measure of one angle or the length of one side and two angles permits you to determine all the lengths and angles of a triangle. Look at https://www.mathsisfun.com/algebra/trig-sine-law.html and https://www.mathsisfun.com/algebra/trig-cosine-law.html. These two laws are what you need for most practical work involving trigonometry.

A decent calculator will let you calculate sines and cosines. Just set the calculator's mode to degrees rather than to radians. (A radian is a measure of angles that is is very useful in calculus, but seldom used in practical work.)

 

[Dr.Peterson]

Is there a formula that can accept a vertex angle and leg length and spit out the base length? Or, better yet, is there a clearer solution I'm not thinking of? Is doing it visually with a tool easier than any of this? I hope I've been clear.
Thanks

It occurs to me that a suitable tool could be the cross-staff, which was used to measure angles visually. It can be easily made and marked off for whatever angles you want to use, then used the same way you would stand at the speaker and spread your arms, but more accurately. Here are two sources I found that tell how to do the calculation (which is very similar to the calculation you were asking for, but easier):

• http://astronomy.swin.edu.au/cosmos/C/Cross-staff
• https://www2.humboldt.edu/scimus/MedInst_rap/X_Staff/Cross-Staff_Maker.htm

The formula in the first finds the angle from the height and base (rather than your leg and base) of an isosceles triangle; you can find the dimensions to use for a given angle by solving for a or b. The second tells how to mark the device for given angles.

 

[JainJude]

Hello,
I am not especially well versed in Geometry vocab, and so I'm having trouble finding the information I need.
I am an audio tech and I'd like an easy way to be more precise when placing my speakers. These speakers have a horizontal coverage expressed in degrees (typically 60-100º), that can be simplified to imagine as a circular arc. It's incredibly important that I be able to cover an area evenly while avoiding spilling into the stage area (this causes feedback, an irritating screeching/droning noise). I use a laser measure to easily mark distances on the job, and would like to be able to essentially draw myself an isosceles triangle that reflects my coverage pattern width.
Let's say that my 60º speaker is my vertex point, and one outer edge of my coverage marks the point of one base angle. From there I need to know my base length, and with that I can make accurate decisions about my coverage instead of standing behind my speaker with my hands splayed in an angle that i think looks correct.xvideos xnxx xxx


Is there a formula that can accept a vertex angle and leg length and spit out the base length? Or, better yet, is there a clearer solution I'm not thinking of? Is doing it visually with a tool easier than any of this? I hope I've been clear.
Thanks

my issue got solved!!!