Find the missing length (trig)

1. Feb 26, 2013

uperkurk

Am I doing this correctly.

$$x \tan=\frac{183}{\tan 30}=317ft$$

Also if it were the other way around and I needed to find the height but I already had the length would it just be

$$h \tan=\frac{x}{\tan 60}=hft$$

2. Feb 26, 2013

jbunniii

I'm not sure what the notation $x \tan$ and $h \tan$ means.

If $h$ refers to the height of the triangle (currently labeled 183 ft), then both of the following are true:
$$\tan(60) = \frac{x}{h}$$
and
$$\tan(30) = \frac{h}{x}$$
So if you know $x$ but not $h$, then you can find $h$ by either
$$h = \frac{x}{\tan(60)}$$
or
$$h = x\tan(30)$$
And if you know $h$ but not $x$, then you can find $x$ by either
$$x = h \tan(60)$$
or
$$x = \frac{h}{\tan(30)}$$

3. Feb 26, 2013

uperkurk

Thanks. I have 1 more question if you wouldn't mind confirming my answer. Just so I don't make another thread.

A ladder is leaning against a wall. The foot of the ladder is 6.25 feet from the wall.
The ladder makes an angle of 74.5° with the level ground. How high on the wall does the ladder
reach? Round the answer to the nearest tenth of a foot.

After working out the remaining angle being 15.5° I then have:

$$x \tan=\frac{6.25}{\tan 15.5}=23ft$$

4. Feb 26, 2013

jbunniii

Yes, it's correct, except again I'm not sure why you wrote "$x \tan$" instead of just $x$.

By the way, homework and homework-like questions should go in the homework forums.

5. Feb 26, 2013

uperkurk

x tan is just the notation they use in this book I'm learning from. It's not really a homework question I'm an independant learner and this book doesn't have the answers in the back so just wanted someone to check I was doing these correctly.

Thanks!

6. Feb 26, 2013

HallsofIvy

Staff Emeritus
Are you sure they are not writing something like "x tan(60)" where "x" is the length of the "opposite side" to get the "near side"?

7. Feb 26, 2013

uperkurk

Nope, the side is just labled as $$x$$ and then it just says

$$x\tan= ...$$