Solving triangle: A train travels 3.5 km on a straight track

[smsmith]

A train travels 3.5 km on a straight track with a grade of 1° 10' (one degree, ten minutes). What is the vertical rise of the train in that distance?

I know that 1° 10' is approx. 1.667 degrees. I can not, however, figure out if I am using the tangent, sine, or cosine in setting the equation. Any advice would be greatly appreciated.

 

[stapel]

Draw the picture: You've got the train going up at an angle, the horizontal distance, and the vertical distance.

The slanty angled distance is "3.5", with the angle being 1¹/₆°. The horizontal distance you don't actually care about (since the question doesn't ask), and the vertical distance is the height "h" of the triangle.

Which ratio uses the angle, and the "opposite" and "hypotenuse"?

Eliz.

 

[smsmith]

That part I get sin(1.667). however, when I set the equation is the problem.
Is it sin(1.667)*3.5=h or am I meesing this up?

My real problem is no matter how I do it none of my solutions match the solution manual.

I have tried everythig and am looking for conformation or denial of the solution of 71.3 meters

 

[smsmith]

OOOh! I get it know. My error was in rounding .
Thank you! You were very helpful.

Is there an easy way to remeber which function I need to use depending on side and angele given?

I struggle with that the most.

 

[stapel]

Is there an easy way to remeber which function I need to use depending on side and angele given?

Probably... but I just draw the picture.

Eliz.