A Ship A Steaming due north at 10km per hour sight at
noon, a second ship B in
direction due east and 1² km
away steaming north east at
SKM per hour - Calculate to the
nearest degrees the bearing
of A from B at 12:15pm?

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°

y∈r, 8> y≥0

hope i !

p.s the first bit only matters if you know what that means.

p.p.s brainliest would really