Question 253576
A mine shaft goes due west 75 m from the opening at an angle of 25° below
the horizontal surface. It then becomes horizontal and turns 30° north of west
and continues for another 45 m. What is the displacement of the end of the
tunnel from the opening. 
-------------
If you know how to graph in 3D,
the start point is (0,0,0)
the turning point is (-75cos(25), 0 ,-75sin(25))
the final position is (-75cos(25) ,-45,-75sin(25))
-------------------------------
Distance from start to finish:
d = sqrt((75cos(25))^2 + (45)^2 + (75sin(25))^2)
d = sqrt(5526.47+2025+98.53)
d = 87.46 meters
==========================
Cheers,
Stan H.