Question

The boundaries PQ, QR, RS and SP of a ranch are straight lines such that: Q is 16km on a bearing of 040° from P, R is directly south of Q and east of P and S is 12km on a bearing of 120° from R.

(i) The distance, in kilometers, of P from S.

(ii) The bearing of P from S.

(c) Calculate the area of the ranch PQRS in square kilometers.

Answer 1

The sketch is quite easy to make.
In mine, the property is made up of a right-angled triangle PQR with base angle of 50° and a hypotenuse of 16 plus an obtuse angled triangle PRS with
RS = 12 and angle PRS = 150°

i)
find PR: cos 50° = PR/16
PR = 16cos50° = ....

now to the other triangle, using the cosine law:
PS^2 = PR^2 + 12^2 - 2(12)(PR)cos150°
you know PR from the first part, then you can find PS

ii) sin(angle RPS)/12 = sin 150°/PS , you found PS in i)
use your calculator to find the angle

iii) add the two triangles, the first one is easy it is right-angled
the second one:
area = (1/2)(PR)(12)sin150° = ....

Answer 2

PQR is right triangle
<QPR = 90-40 = 50 deg
sin 50 = QR/16 so QR = 12.26 km
cos 50 = PR/16 so PR = 10.28 km
area of PQR triangle = (1/2)(10.28 *12.26) = 63.04 km^2
now work on triangle PRS
angle PRS = 180 - 30 = 150 deg
so
PS^2 = 10.28^2 + 12^2 = 2 * 10.28 * 12 cos 150
= 105.7 + 144 - 246.7 cos 150
= 249.7 - 246.7 * -.866 = 249.7+213.6 = 463.3
so PS = 21.5 km
for angle RPS law of sines
sin 150/21.5 = sin RPS/21.5
solve for angle RPS
then compass bearing P to S is 90 + RPS clockwise from North
add 180 deg to get compass bearing angle from S to P
I will leave to you to find area of triangle PRS now and add it to 63.04 to get total area of ranch