Find the missing length (trig)

[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

 

[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
$$x = \frac{h}{\tan(30)}$$

 

[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

 

[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.

 

[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!

 

[HallsofIvy]

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"?

 

[uperkurk]

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

x\tan= ...