The bearing from C to A of the given travel path is gotten as; 100.63°
From the bearing image attached, we can use trigonometric ratios to get;
tanx = opp/adj
x = 53 + θ ---------- eqn(1)
opp = 119
adj = 59
Thus;
tan x = 119/59
tan x = 2.0169
x = tan⁻¹(2.0169)
x = 63.63°
From eqn (1), we have;
θ = x - 53
θ = 63.63 - 53
θ = 10.63°
Thus, the bearing from C to A = 90 + 10.63 = 100.63°
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