Can a man see the three volcanic peaks from Oahu?

[aoiluna]

This is a huge trig problem to find out if a man can see these three Hawaiian volcanoes on different islands from Oahu.

The heights of the three island peaks vary, but they all look the same height from the southeast corner of Oahu. Lanai is 3370 feet above sea level, Maui's Haleakala is 10023 feet above sea level, and Mauna Kea on the big island is 13796 feet above sea level.

The distances from Oahu to the three islands are as follows: to Lanai, 65 miles,to Maui, 110 miles, and to the big island is 190 miles. This is the length of the arc from Oahu to directly below each peak at sea level.

I am to figure out if it is possible to see these three peaks from Oahu. I had to find out the circumference of the earth based on the radius being 3960 miles, and found it to be 24,881.41 miles.

Then I am to find the ratio of the distance to the islands in relation to the circumference of the earth. The paper says that I need to consider that the tourist standing on the shore taking the picture would have a line of sight tangent to the surface of the earth at that point, and then I need to find the angle formed at the center of the earth.

I then should find the length of the hypotenuese and then decide if Mauna Kea is visible from Oahu, and repeat with the other two.

Question 6 asks if this one volcano Kamakou on the island Molokai is visible from Oahu. Molokai is 40 miles from Oahu, and the volcano is 4961 feet above sea level.

Finally, I am to name the three volcanoes visible from Oahu.

This problem is giving me a headache, ecspecially because trig is not my thing. Any help at all would be SUPER appreciated. Thanks!

 

[stapel]

...trig is not my thing.

But surely you can make some progress...?

Eliz.

P.S. to tutors: I think the seven questions in play are the ones listed here.

 

[aoiluna]

I drew a picture..

Well I did draw a picture, and found out that the tangent of alpha is 190/3960. This is for the first mountain, Kea. The cos is 3960/ 3960+H(height of mountain) and the sin is 190/3960+H. 190 is the tourist's line of sight, and the radius is 3960. now im just having trouble finding out the angles.

 

[JakeD]

Re: I drew a picture..

Well I did draw a picture, and found out that the tangent of alpha is 190/3960. This is for the first mountain, Kea. The cos is 3960/ 3960+H(height of mountain) and the sin is 190/3960+H. 190 is the tourist's line of sight, and the radius is 3960. now im just having trouble finding out the angles.

Until Soroban comes along and draws one of his remarkable graphs, I'll say this. The tourist's line of sight is NOT 190. 190 is the length of the arc from Oahu to the base of Kea. To get the angle alpha in radians, divide 190 by the circumference of the earth and then multiply by 2pi. Then solve the equation cos(alpha) = 3960/(3960+H) for H, which is the minimum height Kea must have for it to be visible.

 

[aoiluna]

Thanks! I dont really understand why multiplying it by 2pi will give me the angle in radians, but im taking yuour word. So I divided 190 by the C of the earth and got .00763 and then multiplyed it by 2pi and got .0479, which should be angle alpha in radians, right? I plugged that into your equation and got 4.56, which is the minimum heigt Kea must be to be visible? i think im doing something wrong.

 

[JakeD]

Thanks! I dont really understand why multiplying it by 2pi will give me the angle in radians, but im taking yuour word.

How many radians is 360 degrees?

So I divided 190 by the C of the earth and got .00763 and then multiplyed it by 2pi and got .0479, which should be angle alpha in radians, right? I plugged that into your equation and got 4.56, which is the minimum heigt Kea must be to be visible? i think im doing something wrong.

4.56 is correct. What does it mean? What is the unit?

 

[aoiluna]

According to that it is 4.56 feet above sea level. and this would be the minimum height that Mount Kea would have to be to see it form Oahu?