# Thread: Trig Problem Solving question

1. ## Trig Problem Solving question

In a movie cinema the screen is 2.5 m high and 3 m off the ground. if the ground is flat and the average eye level height for the audience is 1.2 m: how do you work out the angle from the persons eyes to the tops and bottom of the screen.

and: what is the prime distance away from the screen to get the maximum angle of screen vision.

sorry if it is a bit confusing...

2. Originally Posted by iamdylan123
In a movie cinema the screen is 2.5 m high and 3 m off the ground. if the ground is flat and the average eye level height for the audience is 1.2 m: how do you work out the angle from the persons eyes to the tops and bottom of the screen.

and: what is the prime distance away from the screen to get the maximum angle of screen vision.

sorry if it is a bit confusing...
I attached a picture, You can see that the screen and the ground line form 2 right triangles. Now how would you solve for b and t?

(consider the x as a value which is known)

3. thats helpful but i need neither angles b or t i need the angle from the top of the screen to the bottom of the screen so its an angle of elevation not a right angle triangle? if that makes sense?

Call this new angle theta with two lines coming from the eyeball. one going to the top of the screen and the other going to the bottom.

EDIT: Sorry about the double post

EDIT EDIT: Tan b=1.8/x
tan t=2.5/x

4. Originally Posted by iamdylan123
thats helpful but i need neither angles b or t i need the angle from the top of the screen to the bottom of the screen so its an angle of elevation not a right angle triangle? if that makes sense?

Call this new angle theta with two lines coming from the eyeball. one going to the top of the screen and the other going to the bottom.

EDIT: Sorry about the double post
b and t are called for in your question: "how do you work out the angle from the persons eyes to the tops and bottom of the screen."

To find theta, $\theta = t - b$

consider a clock pointing to the 10, and you want to know how many digits it must go through to get to the 12. Well, 12 - 10 = 2. This is the same thing, take the top angle (t) and subtract the bottom angle ( b) and you will have the angle which is the difference, the one you have labeled theta.

5. thank you so much! so now theta = t-b

and t and b are both:
tan t=2.5/x
tan b=1.8/x

So that must mean the greater x is then the greater theta will be?

EDIT: but that doesnt make sense? beacuase the text book calls for the distance to the nearest 10cm that will give the viewer maximum viewing angle

6. Originally Posted by iamdylan123
thank you so much! so now theta = t-b

and t and b are both:
tan t=2.5/x
tan b=1.8/x

So that must mean the greater x is then the greater theta will be?

EDIT: but that doesnt make sense? beacuase the text book calls for the distance to the nearest 10cm that will give the viewer maximum viewing angle
First of all: I wrote the wrong number in the image, sorry, I updated it (should have been 3-1.2 not 3-1.8)

For the top, the screen is 2.5 meters tall, but it is 3 meters off the ground, so it is 2.5+3 = 5.5 meters. But we moved the whole graph down (notice the eyeball is at the origin instead of 1.2 meters above the origin) So the top is 5.5 - 1.2 = 4.3 meters

The bottom starts at 3 meters off the ground, but our ground is 1.2 meters below our origin (because we moved our eyeball down to the origin) so the bottom is 3-1.2 = 1.8 meters above the ground.

Now you can use those values in your equation.

For the other part, I think you are confused. As x decreases, the angle decreases, because both angles are moving towards being directly above the eye (think of sitting on the front seat in a movie theatre, and how you are staring almost straight up. As x increases, the screen moves back, and so both angles are going down towards the ground and thus decreasing (this is why objects seem smaller as you get farther away from them).

So somewhere in the middle is our "sweet spot" where theta is at it's largest.

7. I know that the top of the screen is 5.5 m from the ground and the bottom of the screen is 3 m from the ground.

and i know that 5.5-1.2=4.3 which is the top of the screen of the right angled triangle

and 3m-1.2=1.8 which is the bottom of the screen from the right angled triangle

but what i need to know is that if x and theta are in direct proportion, because if they are the question contradicts itself which is easy to explain, but if they are not in direct proportion, how would you work out the ideal distance away from the screen (x) to have the largest viewing angle (theta).

let x=1

t= tan t=4.3/1
tan t=0.0751
tan^-1 0.0751 = 4.29 degrees

b= tan b=1.8/1
tan b=0.0314
tan^-1 0.0314=1.79 degrees

So Theta = t-b
theta = 2.5 degrees?

is that working correct?
tan b

8. $tan(t) = \frac{4.3}x$

$t = \arctan\left( \frac{4.3}x \right)$

$tan(b) = \frac{1.8}x$

$b = \arctan\left( \frac{1.8}x \right)$

$\theta = t - b$

$\theta = \arctan\left( \frac{4.3}x \right) - \arctan\left( \frac{1.8}x \right)$

I know how to solve the maximum value with Calculus, but not with Trig :/

I'm getting $\sqrt{7.74}$ meters for x with Calculus.

Can someone look at this and see if there is a better way / if I have assessed it incorrectly? It seems unlikely that a Trig problem would be this involved.

9. So do you say that if you are only using trig functions. Substituting x for one then 2 then 3 etc. would be the only way to work it out?

10. Originally Posted by iamdylan123
So do you say that if you are only using trig functions. Substituting x for one then 2 then 3 etc. would be the only way to work it out?
Better plan would be to graph it and then approximate it. But it is more likely that either I made an error somewhere, or I am overlooking a Trig method that I should be seeing, or I have done the whole problem wrong, or the people who wrote your book are sadists.

We should wait for someone else to check my work, I'm not very confident in it. It seems good when I look at it, but the answer was much more complicated that it seems like it should have been (I could give you a basic explanation of how I got it, but I'm not sure if you'd understand it, and if you got the concept, you still wouldn't understand the steps).

11. Okay, I think I found where our problem is: They ask "what is the prime distance away from the screen to get the maximum angle of screen vision."

I googled the term "prime distance" and got Prime Distance -- from Wolfram MathWorld

"The prime distance pd(n) of a nonnegative integer n is the absolute difference between n and the nearest prime."

So in this case, the square root of 7.74 is about 2.78, which is pretty close to the prime number 3. So lets say that the answer they are looking for is the number 3.

If you had to do it yourself, you could find it by plugging in successive prime numbers into the equation and going with the highest. ie, try 2, 3, 5, 7, 11. (or graph it and see which is closest)

Here is a picture of the graph. The pink line is the line of the actual maximum at the square root of 7.74, the green lines represent the prime numbers. You can see that the maximum value is closest to three, and since we are looking for the prime distance from the origin to the maximum, we choose 3 as it is closest to sqrt(7.74)

12. Thanks for your help. Sadism is a real problem down here in australia. That is likely to be the problem...

Also, it might be a bit suspicious if a grade 11 student suddenly knows calculus haha.

Graphing sounds good but then again, i dont have very much time in the exam.

It seems very hard and time consuming for an exam thought doesn't it?

They copied the question straight out of the text book, the lazy public school teachers probably didn't even look at it properly.

13. Originally Posted by iamdylan123
Thanks for your help. Sadism is a real problem down here in australia. That is likely to be the problem...

Also, it might be a bit suspicious if a grade 11 student suddenly knows calculus haha.

Graphing sounds good but then again, i dont have very much time in the exam.

It seems very hard and time consuming for an exam thought doesn't it?

They copied the question straight out of the text book, the lazy public school teachers probably didn't even look at it properly.
I edited the post to include what I think is the answer, don't know if you saw it.

14. Yeah i did, thanks for your help, it helped alot, but it i just write down the answer, i fail that question, but if i write down some of the stuff that we found out i can pass with a high C i think.

But if you think of a simpler trig way that you overlooked let me know, but otherwise you have helped a lot thank you.

15. ## Re: Trig Problem Solving question

thx the teacher didn't change the question for the assignment

Page 1 of 2 12 Last