[SOLVED] Law of Sines/Cosine Word Problems

• May 23rd 2006, 03:25 PM
robgoatse
[SOLVED] Law of Sines/Cosine Word Problems
Need to study these for a test, but I wasn't there when she gave us the solutions. Can anyone help? Thanks in advance.

1) You need to determine the length of a tunnel drilled through a lunar peak at the center of a crater. From a point on the crater floor, you run ropes to the ends of the tunnel. The first rope's distance covered is 24.5 meters and the second rope's distance is 21.2 meters. The two ropes meet at an angle of 68 degrees.
a) What is the length of the tunnel?
b) At what angle with the 1st rope should the tunnel be drilled so that it comes out at the correct exit point?

2) The road from Westville to Centerville is 10 miles long, and from Centerville to Eastville is 15 miles long. The two roads meet at an angle of 126 degrees. A telephone company wants to place a cable that connects Westville and Eastville directly.
a) How long will the cable be?
b) What angle will the cable make with the road from Eastville to Centerville?

3) The Daredevil cliffs rise vertically from the beach. The beach slopes gently at an angle of 3 feet from the horizontal. Laying at the water's edge, 50 feet from the base of the cliff, you determine your position, and a straight line to the top of the cliff creates an angle of 70 degrees with the line to the base of the cliff.
a) How high is the daredevil cliff?

4) A guide needs to know the elevation gain from the floor of Yosemite Valley to the top of El Capitan, a cliff on one side of the valley. She measures 3300 feet between two points on the valley floor next to the cliff and sights the top of El Capitan at the angles shown. (ATTACHED)
a) Find distance AC in the diagram
b) Use the answer party (a) to find the height of point C above the valley floor.

5) A crew is given the job of measuring the height of a mountain. From a point on level ground, they measure and angle of elevation to the top of 21 degrees. They move 507 meters closer and the angle is now 35 degrees.
a) How high is the mountain?

6) You observe two hot air balloons at equal altitudes. The 1st balloon is 12 miles from the airport and the second balloon is 17 miles from the airport. The observed angle between them is 70 degrees.
a) How far apart are the balloons?
b) What angle will the first balloon make with the straight line to the airport and the straight line to the 2nd balloon?

I really appreciate any help.
• May 23rd 2006, 07:59 PM
earboth
problem nr. 2 only
Quote:

Originally Posted by robgoatse
Need to study these for a test, but I wasn't there when she gave us the solutions. Can anyone help? Thanks in advance.
...
2) The road from Westville to Centerville is 10 miles long, and from Centerville to Eastville is 15 miles long. The two roads meet at an angle of 126 degrees. A telephone company wants to place a cable that connects Westville and Eastville directly.
a) How long will the cable be?
b) What angle will the cable make with the road from Eastville to Centerville?
...

Hello,

use the law of Cosine.

to a) You know all the values necessary to calculate the distance WE:
$d(WE)=\sqrt{10^2+15^2-2 \cdot 10 \cdot15 \cdot \cos(126^\circ)}$ $\approx 22.39 miles$

to b) Transform the law of Cosine so that it expresses the Cosine of an angle and then plug in all values you know:
$\cos(\epsilon)=\frac{10^2-15^2-22.39^2}{-2 \cdot 15 \cdot 22.39}\approx .93244...$
Now use the \arccos()-function and you'll get an angle 21.18°.

Greetings

EB
• May 24th 2006, 07:40 AM
earboth
problem nr. 5 only
Quote:

Originally Posted by robgoatse
Need to study these for a test, but I wasn't there when she gave us the solutions. Can anyone help? Thanks in advance.
...

5) A crew is given the job of measuring the height of a mountain. From a point on level ground, they measure and angle of elevation to the top of 21 degrees. They move 507 meters closer and the angle is now 35 degrees.
a) How high is the mountain?
...
I really appreciate any help.

Hello,

I've attached a diagram to show you what i calculated.

You deal with 2 right triangles. Right right triangle:
$\tan(35^\circ)={h\over x}\Longrightarrow x={h\over \tan(35^\circ)}$

Left right triangle:

$\tan(21^\circ)={h\over (x+507)}$ Substitute the x by the expression of the first equation and you'll get:

$\tan(21^\circ)={h\over ({h\over \tan(35^\circ)}+507)}$

After a few transformations (using blood, sweat and tears) you'll get
$h \approx 430.78 m$

Greetings

EB
• May 24th 2006, 09:50 PM
earboth
problem nr. 6 only
Quote:

Originally Posted by robgoatse
Need to study these for a test, but I wasn't there when she gave us the solutions. Can anyone help? Thanks in advance.
...
6) You observe two hot air balloons at equal altitudes. The 1st balloon is 12 miles from the airport and the second balloon is 17 miles from the airport. The observed angle between them is 70 degrees.
a) How far apart are the balloons?
b) What angle will the first balloon make with the straight line to the airport and the straight line to the 2nd balloon?
I really appreciate any help.

Hello,

I've attached a diagram to show the situation.

to a)
Use the law of cosine: (d(B1B2) means the distance between the two balloons)

$d(B1B2)=\sqrt{12^2+17^2-2 \cdot 12 \cdot17 \cdot \cos(70^{\circ})}\approx 17.13 miles$

to b)
Here it doesn't matter which law you use. But as I presume you have stored the resultof the distance it will easiest to use the law of cosine again. Alpha means the angle at the airport: (B1-A-B2):

$\cos(\alpha)={{17.13^2-12^2-17^2}\over {-2\cdot 12 \cdot 17}} \approx .3420$
Use the arccos-function and you'll get 70°. That means the triangle B1AB2 is (nearly) isosceles.

Greetings

EB