Question

A skier moves 100.0 m horizontally, and then travels another 35.0 m uphill at an angle of 35.0º above the horizontal. What is the skier's displacement from his starting point? (Use the graphical method to answer and include your scale on the graph paper.) Give both the magnitude and the angle in your answer.

Answer 1

The graph is in the attachment.
x = 35 · cos 35° = 35 · 0.819 = 28.67 m
y = 35 · sin 35° = 35 · 0.574 = 20.075 m
d ² = 128.67² + 20.075²
d ² = 16,958.97425
d = 130.234 m 
cos α = 128.67 / 130.23 = 0.988
α = 8.88° ≈ 8° 53 `

Answer 2

the vectors make a triangle where: 
The base AB is 100m 
The slope BC is 35m at 35.0º to the horizontal : 
so the height (h) of C above AB = 35 * sine(35.0º) 
The horizontal distance from B = 35 * cosine(35.0º) 
The horizontal distance from A = 100 + 35 * cosine(35.0º) 

Tan (theta) = 35 * sine(35.0º) / (100 + 35 * cosine(35.0º))