# Thread: Find H on this triangle

1. ## Find H on this triangle

here's the pic I made... (find H)

now, I'm confused on how to set it up, should it be

$tan(18) = h / 5+x$
$tan(32) = h / x$

or

$tan(18) = h / b$
$tan(32) = h / b-5$

(b = base)

or neither?

please tell me which set-up it is and why, i can't get it

2. Originally Posted by Arez
here's the pic I made... (find H)

now, I'm confused on how to set it up, should it be

$tan(18) = h / 5+x$
$tan(32) = h / x$

or

$tan(18) = h / b$
$tan(32) = h / b-5$

(b = base)

or neither?

please tell me which set-up it is and why, i can't get it
Both are correct, except that you should use parentheses for the denominators so that
h/ 5+x would appear h / (5+x)
h / b -5 would appear h /(b-5)

tan is opposite side / adjacent side .......that's it.
No matter how you come up with the measurements of the sides, as long as you carefully follow tan = opp/adj, then you're okay.

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EDIT:

Opsss, I'm wrong!
After reading masters' reply (which is correct), I am wrong with mine.
tan(32 deg) cannot be h /(5 +x) nor h /(b -5)....because h is not the opposite side of 32 degrees.

Must be that I just woke up.
Still, I shouldn't have missed that tan(32 deg)....

Sometimes, though, I like being wrong. :-)

3. Originally Posted by Arez
here's the pic I made... (find H)

now, I'm confused on how to set it up, should it be

$tan(18) = h / 5+x$
$tan(32) = h / x$

or

$tan(18) = h / b$
$tan(32) = h / b-5$

(b = base)

or neither?

please tell me which set-up it is and why, i can't get it
Not enough info, I fear. According to your diagram:

$\tan 18=\frac{H}{5+X}$ is correct!

The 32 degree angle has no opposite side defined. H is the opposite side of the 18 degree angle. So,

$\tan 32 \neq \frac{H}{X}$

4. Originally Posted by ticbol
Opsss, I'm wrong!
After reading masters' reply (which is correct), I am wrong with mine.

Sometimes, though, I like being wrong. :-)
It keeps us humble, you know.