# 16:9 n inch diag == width/height? **URL:** \ **Category:** Factual Questions\ **Created:** [April 11, 2003, 6:08am UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988 "2003-04-11T06:08:41Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![squeegee](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/squeegee/32/14537_2.png) [@squeegee](https://boards.straightdope.com/u/squeegee)\ **Post date:** [April 11, 2003, 6:08am UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/1 "2003-04-11T06:08:41Z")
If a TV has 16:9 aspect, and the diagonal size is n inches, what is the width/height of the picture? It’s late, I’ve been fiddling in Excel for 30 minutes on this, and it’s really bugging me. I keep getting something like sqrt((diag\*diag) \* hiAspect )/loAspect. Not. !@#!@ing Pythagoras 🙂 ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [April 11, 2003, 6:27am UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/2 "2003-04-11T06:27:05Z")
h=n(sin(arctan(9/6))) w=n(cos(arctan(9/16))) [http://www.jjgifford.com/expressions/geometry/trigonometry.html](http://www.jjgifford.com/expressions/geometry/trigonometry.html) or use h^2+w^2=n^2 and h/w=9/16 w=n/sqrt[(9/16)^2+1] h=n/sqrt[(16/9)^2+1] check my math, but the methodology is correct ---
**Author:** ![Happy\_Fun\_Ball](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/happy_fun_ball/32/17664_2.png) [@Happy\_Fun\_Ball](https://boards.straightdope.com/u/Happy_Fun_Ball)\ **Post date:** [April 11, 2003, 6:27am UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/3 "2003-04-11T06:27:17Z")
So, if what you are saying is that the height is 9/16 of the width, then: w= 16n (337)[sup]-1/2[/sup] h=9n (337)[sup]-1/2[/sup] ---
**Author:** ![Happy\_Fun\_Ball](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/happy_fun_ball/32/17664_2.png) [@Happy\_Fun\_Ball](https://boards.straightdope.com/u/Happy_Fun_Ball)\ **Post date:** [April 11, 2003, 6:28am UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/4 "2003-04-11T06:28:52Z")
Which is what John Mace Said… Next time I’ll preview… ---
**Author:** ![Linus\_Van\_Pelt](https://avatars.discourse-cdn.com/v4/letter/l/ebca7d/32.png) [@Linus\_Van\_Pelt](https://boards.straightdope.com/u/Linus_Van_Pelt)\ **Post date:** [April 11, 2003, 7:49am UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/5 "2003-04-11T07:49:48Z")
I hope I’m not presuming to translate for the slightly less geekily inclined. For the height: square the diagonal, divide by 4.16, then take the square root of the whole thing. For the width: square the diagonal, divide by 1.32, then take the square root of the whole thing. It’s a close approximation using decimal conversions of the fractions, but it’s a lot easier with a basic pocket calculator and it comes out within a very small fraction of an inch. ---
**Author:** ![squeegee](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/squeegee/32/14537_2.png) [@squeegee](https://boards.straightdope.com/u/squeegee)\ **Post date:** [April 11, 2003, 2:06pm UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/6 "2003-04-11T14:06:43Z")
Thanks, gang. I think John Mace’s second example was what I was going for, but thanks for the different approaches. ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [April 11, 2003, 4:44pm UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/7 "2003-04-11T16:44:15Z")
Talk about making things unnecessarily complicated. If the diagonal is n and the ratio is 16:9 then the sides are 0.87_n and 0.49_n ---
**Author:** ![Happy\_Fun\_Ball](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/happy_fun_ball/32/17664_2.png) [@Happy\_Fun\_Ball](https://boards.straightdope.com/u/Happy_Fun_Ball)\ **Post date:** [April 11, 2003, 5:50pm UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/8 "2003-04-11T17:50:39Z")
> [@](#): > > _Originally posted by sailor \* > \*\*Talk about making things unnecessarily complicated. If the diagonal is n and the ratio is 16:9 then the sides are 0.87_n and 0.49\*n \*\* That’s what I said… I just like having all the decimal places. Knowing the measurements to 1/10000 of an inch is important when trying to decide where to hang your plasma TV. 😉 ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [April 11, 2003, 6:01pm UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/9 "2003-04-11T18:01:19Z")
\>\> Knowing the measurements to 1/10000 of an inch is important when trying to decide where to hang your plasma TV. So why are you neglecting to take into account air temperature, pressure and humidity? 😉 ---
**Author:** ![squeegee](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/squeegee/32/14537_2.png) [@squeegee](https://boards.straightdope.com/u/squeegee)\ **Post date:** [April 12, 2003, 6:01am UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/10 "2003-04-12T06:01:28Z")
> [@](#): > > _Originally posted by sailor \* > \*\*Talk about making things unnecessarily complicated. If the diagonal is n and the ratio is 16:9 then the sides are 0.87_n and 0.49\*n \*\* How do you derive that? ---
**Author:** ![bibliophage](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bibliophage/32/7615_2.png) [@bibliophage](https://boards.straightdope.com/u/bibliophage)\ **Post date:** [April 12, 2003, 1:08pm UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/11 "2003-04-12T13:08:36Z")
Let the sides be 16x and 9x and the diagonal be n. Use the Pythagorean theorem and solve for x in terms of n. Multiply x by 9 to get one side and by 16 to get the other. Since 337 is the sum of 9 squared and 16 squared, 0.49n = 9n/sqrt(337) 0.87n=16n/sqrt(337) ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [April 12, 2003, 1:55pm UTC](https://boards.straightdope.com/t/16-9-n-inch-diag-width-height/167988/12 "2003-04-12T13:55:25Z")
Or, looking at it another way: if one side of the rectangle is 16 and the other one is 9, then the diagonal is 18.36. Any screen with the same proportion will have proportional dimensions: ```auto n a b ------- = ------- = ------- 18.36 16 9 ``` From where: a = n \* 16 /18.36 b = n \* 9/18.36