Need Help With 2 Word Problems

Printable View

April 24th 2006, 12:46 PM
wkid87

Need Help With 2 Word Problems and Checking 2 Problems

The biggest bridge in the world is the bridge over the Royal Gorge of the Arkansas River in Colorado. Sightings to the same point at water level directly under the bridge are taken from each side of the 880-foot long bridge at angles of 69.2 degrees and 65.5 degrees. How high is the bridge?

So what I did was made one big triangle and split in up in half, where i have a right triangle but 2 triangle inside of 1. Will the adjacent be 90 degrees? I also, have the opposite angles at 69.2 on the 1st triangle and put 65.5 on the 2nd triangle. I tried to work it out from there but it didn't work. What would I have to do with the 880 foot? Split it?

Suppose we have an oblique triangle such that tan of alpha=1 and tan of beta=x. Find tan of gamma in terms of x.

I'm tottally lost on that problem. Thanks!
April 24th 2006, 08:46 PM
earboth

1 Attachment(s)

Quote:

Originally Posted by wkid87

The biggest bridge in the world is the bridge over the Royal Gorge of the Arkansas River in Colorado. Sightings to the same point at water level directly under the bridge are taken from each side of the 880-foot long bridge at angles of 69.2 degrees and 65.5 degrees. How high is the bridge?
So what I did was made one big triangle and split in up in half, where i have a right triangle but 2 triangle inside of 1. Will the adjacent be 90 degrees? I also, have the opposite angles at 69.2 on the 1st triangle and put 65.5 on the 2nd triangle. I tried to work it out from there but it didn't work. What would I have to do with the 880 foot? Split it?...

I'm tottally lost on that problem. Thanks!

Hello,

to your 1rst problem:
All what you've suggested is OK with your problem:
I've attached a diagram to demonstrate what I've calculated.

1. Split the length of 880' int x and (880'-x). Then you have 2 right triangles, with which you get 2 equations:
$\frac{x}{h}=\tan(69.2^\circ)$ and $\frac{880-x}{h}=\tan(65.5^\circ)$

From the 1rst equation you get: x = h * tan(69.2°). This result plug into the 2nd equation:

$\frac{880-h \cdot tan(69.2^\circ)}{h}=\tan(65.5^\circ)$

Expand the LHS of this equation to:

$\frac{880}{h}- tan(69.2^\circ)=\tan(65.5^\circ)$
Solve for h and you'll get:

$h=\frac{880}{\tan(65.5^\circ)+\tan(69.2^\circ)} \approx 182.3'$

And now I've to run to catch some €.

Greetings

EB
April 25th 2006, 09:16 AM
wkid87

Thanks, so much. Now I was able to understand it better. Now the task of figuring out the unknown in the 2nd problem.
April 25th 2006, 08:59 PM
wkid87

Need to check 2 problems!

Just want someone to check these two problems to see if I'm right or wrong, and if I'm wrong what can I do right.

1.According to Little League baseball offfical regulations, the diamond is a
square 60 feet on a side. The pitching rubber is located 46 feet from home plate
on a line joining home plate and second base. How far is it from the pitching
rubber to first base?

Will it be 38.5. I got that from the phatengom(sp) therom.

2.sin (4 theta)- sin (6 theta)=0 and the interval is from 0 to 2 pie.

Will the gerenral solution be? 10 pie/2

Thanks