# Thread: Help with trigonometry world problem.

1. ## Help with trigonometry world problem.

Hi guys,

I am having trouble with this word problem:

Two buildings face each other on opposite sides of a city square. The angle of depression from the top of the taller building to the base of the shorter is 38 degrees. The angle of depression from the top of the shorter building to the base of the taller is 23 degrees. If one building is 30 m higher than the other, how wide is the square?
If someone could point me in the right direction it'd be much appreciated! I've just been banging my head against a wall on this one .

Thanks,

2. Originally Posted by flyingllama
Hi guys,

I am having trouble with this word problem:

If someone could point me in the right direction it'd be much appreciated! I've just been banging my head against a wall on this one .

Thanks,
Here's how I would do it:

Draw a basic diagram of two buildings, on of height $h$, and one of height $h+30$, then you can draw on the two lines of depression, complete with their angles of depression.

Using the alternate angles rule, you then have two right angles triangles. Both of these have the same base - (the width of the square).

Using trigonometry, you end up with the two equations:

$\tan{23} = \frac{h}{x}$ and $\tan{38} = \frac{h + 30}{x}$ where x is the width of the square.

If we rearrange this to make h the subject, we get the following:

$h = x\tan{23}$ and $h = x\tan{38} - 30$

Equating these two eliminates h, and gives you the following:

$x\tan{23} = x\tan{38} - 30$

You should be able to solve this by rearranging and taking out the common factor of x, giving you the width of the square.

Hopefully this wasn't too confusing, I will try and do a diagram to help explain the triangles that we found.

Craig

3. Here's a basic diagram to illustrate what I mean.

4. Hi Craig,

Thanks for your help but I'm still having trouble.

As you stated the equation is:

$
x\tan{23} = x\tan{38} - 30
$

I need to isolate $x$ or remove it from the equation completely?

If I move the second expression over I get:

$
x\tan{23} - x\tan{38} = - 30
$

Then I can remove the x's?

or do I have to isolate the x's?

In that case:

$
\frac{x\tan{23}}{\tan{38}} = x\ - 30
$

Where do I go from here? How do I get the x in the top left out? What am I supposed to be doing?

Apologies for my inability to comprehend these mathematical topics.

The answer in the BOB is 84.1m by the way.

Thanks.

5. Originally Posted by flyingllama
I need to isolate $x$ or remove it from the equation completely?

If I move the second expression over I get:

$
x\tan{23} - x\tan{38} = - 30
$
As x is the width of the square you want to isolate it, not remove.

Notice that in your above equation, everything on the left hand side has a common factor, x, removing this gives you:

$
x(\tan{23} - \tan{38}) = - 30
$

Dividing both sides by $(\tan{23} - \tan{38})$ gives you:

$x = \frac{-30}{(\tan{23} - \tan{38})}$

Now all you have to do is work out what numerical value this is, and there's you answer

The maths in these "word" problems is often not that hard, the difficult thing is translating the words and finding a solvable formula from them.

Anyway hope this helped

Craig

6. Thanks Craig! You are a big help for me and a valuable asset to this forum.