Christopher's distance from his starting point is 3.6 km
Since Christopher initially walks South 5 km and then walks on a bearing of of 036º until he is due east of his starting point, His distance South, his distance from his starting point and his distance from his 036º bearing, all form a right-angled triangle with opposite side to the angle 036º, as his distance from his starting point, x and the adjacent side to the angle 036º, as his distance 5 km south.
Since we have both the opposite and adjacent sides of a right-angled triangle,
From trigonometric ratios,
tanФ = opposite/adjacent
tanФ = x/5 km
Now Ф = 036º
So, tan36º = x/5km
x = 5 km(tan36º)
x = 5 km (0.7265)
x = 3.633 km
x ≅ 3.6 km to 1 decimal place.
So, Christopher's distance from his starting point is 3.6 km.
Learn more about bearing here:
brainly.com/question/11684993
