right triangle problems- trigonemtry?
interstate 5 in California enters the San Joaquin Valley through a mountain pass called the Grapevine. The road descends from an altitude of 3000 feet to 500 feet above sea level in a distance of 6 miles. (a mile is 5280 feet)
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Write “0” on the horizontal line. Write 500 at the base of the triangle, and 3000 at the peak. The height of the triangle is 3000 minus 500. The length of the base of the triangle is 6 times 5280.
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Forgive the crudeness of it but if you should get the idea. The left most side is the 2,500 feet of elevation, the crappy looking downslope is 6 mi or 31,680 feet because you’re gonna have to convert one of the values to be the same as the other, doesn’t matter which, just so long as they are the same. So now you have a right angle on the bottom left and two values. I don’t know what you’re trying to solve so I’ll give you some help (I don’t know you’re math expertise so forgive me if this is insulting). Use Pythagorean Theorem to get the bottom length:
a^2 + b^2 = c^2
In this case: 2500^2 + b^2 = 31680^2
6,250,000 + b^2 = 1,003,622,400
b^2 = 997,372,400
b = 31,581.20
That now gives you all the sides. Using the SSS theorem you can now calculate the angle if you need that too. You can also use SAS theorem.
a / sin A = b / sin B = c / sin C
Hope this gets you going..
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