Please answer each one, trig word problems involving law of sines and law of cosines and bearings, please help ASAP

Question

1) Two people are 3.21 miles apart and both spot a hot air balloon. One person observes the balloon at an angle of elevation of 25 degrees and the other observes the balloon at an angle of elevation of 38 degrees. Determine the altitude of the ballon and round your answer to the nearest hundrendth and be sure to label your answer.

2) A runner starts at point A and runs 860 feet due East to point B. The runner then turns and runs at a bearing of S12°E for 175 feet to point C. How far is point C from point A?

Answer 1

Hi Hannah.

 

For number 1, the question is hoping that you realize that despite the change in angle for each observer, the height of the balloon will be the same for both.

 

You can set up two right triangles: one for observer 1, and one for observer 2. They will both have the same height. Since you know the observers are 3.21 miles apart, you can assume the length of one right triangle is "X" and the other length is "3.21 - X". Both triangles will have a height we'll call "h".

 

Observer 1 sees the balloon at an angle of 25 deg. The height of the triangle is h and the length of the triangle is x. So you can say:

 

tan(25) = h/x

 

Observer 2 sees the balloon at an angle of 38 deg. The height of the triangle is h and the length of the triangle is x. So you can say:

 

tan(38) = h/(3.21 - x)

 

You now have two equations, two unknowns. Solve for x in terms of h the first equation.

 

x = h/tan(25)

 

Plug this into your second equation.

 

tan(38) = h/(3.21 - h/tan(25))

 

Solve for h.

 

tan(38)*(3.21 - h/tan(25)) = h

 

3.21*tan(38) - tan(38)/tan(25)*h = h

 

3.21*tan(38) = h + tan(38)/tan(25)*h

 

3.21*tan(38) = h(1 + tan(38)/tan(25))

 

h = 3.21*tan(38)/(1 + tan(38)/tan(25))

 

Don't forget your answer will be in miles.

 

2) A bearing of S 12deg E means that you first point your directional arrow due south, then rotate it eastward (counterclockwise) 12 degrees. If your runner was running due east, and then traveled S 12deg E, it means he made a 102 deg angle with his original path, since south is 90 degrees from east, and when you go another 12 degrees east from due south. So it winds up being 102 degrees.

 

To figure out how far A is from C, you use the law of cosines. You have two legs of a triangle AB and BC, each of length 860 feet and 175 feet, respectively. You have an angle <ABC which measures 102 degrees. So:

 

AC² = AB² + BC² - 2*(AB)(BC)cos(<ABC)

 

AC² = 860² + 175² - 2*(860)(175)*cos(102)

 

AC² = 739600 + 30625 - 301000*cos(102)

 

AC² = 770225 + 62581.4

 

AC² = 832806.4

 

AC = 912.6 feet

 

Hope this helps.