Question

Can I please have help? There is a diagram attached.

An explorer on the moon moves outward from his spaceship (point O) for a distance of 500 metres to A. He then makes a 90° turn and travels 200 metres.
He turns so that his new direction is perpendicular to the line between himself and the spaceship and moves another 200 metres (point B). He is gradually getting further away from his starting point O.
He continues this process until he’s 700m from the spaceship then returns directly to it. (The diagram only shows the first 4 sections of his trip.)



a) Calculate how far he is from O when he makes each turn (ie. at A, B, C, etc).

b) If he comes straight home when he is 700m away, what is the total distance he travels?

c) What is the angle between the line OA and OB?

d) What is the angle formed by his departure path and his return path?

Answer 1

Answer & Working:

Using Pythogoras' Theorem:

a)

• each turn at A √(500)²+(200)² = 538.5m
• each turn at B √(538.5)²+(200)² = 574.4m
• each turn at C √(574.4)²+(200)² = 608.2m

b)

• total distance = 500 + 4(200) + 700 = 2000m

c)

• tan θ = opposite/adjacent

tan θ = 200/500

θ = tan^-1(2/5)

= 21.8°

d)

• OB-OC = tan^-1(200/538.5)

= 20.4°

• OC-OD = tan^-1(200/574.4)

= 19.2°

• OD-r.p = tan^-1(200/608.2)

= 18.2°

d.p --> r.p = 21.8° + 20.4° + 19.2° + 18.2° = 79.6°

(Correct me if i am wrong)