Question

14. A person walks 80 m east, then tums right through
angle 143° walks further 50 m and stops. His
position relative to the starting point is
(1) 50 m, 53° east of south
(2) 50 m, 53° south of east
(3) 30 m, 37° south of east
(4) 30 m, 53° south of east​

Answer 1

Let A be the starting point and walking 80m east upto C. Then turns 143° right and walk 15m to reach B. So form triangle ABC. Drop a perpendicular from B to AC at E.

<ACB = (180 - 143) = 37°

From triangle BCE,

EC = 15 * cos(37) = 11.979m

EB = 15 * sin(37) = 9.027m

AE = AC - EC = 80 - 11.979 = 68.021m

For triangle AEB,

tan (<BAE) = EB / AE = 9.027/68.021

So <BAE = 7.56°

sin ( 7.56) = BE/AB = 9.027/AB

So, AB = 68.613m

Hence the person is 7.56° south-east from starting point A at a distance 68.613m.

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