still posting this, with no input of your own?
using the law of cosines,
RQ^2 = 240^2 + 120^2 - 2*240*120*cos130°
RQ = 330.19
To find the bearing of R from Q, the displacement is
y = 240 sin40° = 154.26
x = 240 cos40° + 120 = 303.85
So, QR make an angle of
arctan(y/x) = 27° with the x-axis.
So, the bearing of R from Q is 90-27 = 63°
Since PQ = 240, the line from the top of post P to the top of post Q is
d = 240/cos9° = 243
its speed is 243/18 = 13.5 m/s
Double-check my math, and next time, how about including some of your work? You can't be totally clueless ...
Three points P, Q and R are on level ground. Q is 240 m from P on bearing of 230°. R is 120 m to the east of P.
(b) Determine
(i) The distance R from Q;
(ii) The bearing of R form Q.
(c) A vertical post stands at P and another one at Q. A bird takes 18 seconds to fly directly from the top of the post at Q to the pot of the post at P. Given that the angle of depression of the top of the post at P from the top of post at Q is 9°, calculate:
(i) The distance to the nearest meter, the bird covers;
(ii) The speed of the bird in km/h
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