Christopher walks 5km south then walks on a bearing of 036º until he is due east of his starting point. How far is he from his starting point, to 1 decimal place?

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.

This 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.

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