Alliance Form 3 P1 Q19 — Bearings & Scale Drawing

KCSE (Form 3) Form 3 Geometry

Published

The Question

“From a point P, the point Q is 16 km away on a bearing of 040 degrees. R is due south of Q and due east of P. The point S is 12 km from R on a bearing of 120 degrees. (a) Make an accurate scale drawing of P, Q, R and S. (b) Find the distance PS and the bearing of P from S. (c) Find the area of the ranch PQRS.”

1

Set up the positions and the right angle at R

Draw a north line at P and place Q 16 km away on a bearing of 40 degrees. Because R is due south of Q and due east of P, the line PR runs straight east and the line QR runs straight down, so they meet at a right angle at R. Triangle PQR is therefore right-angled at R, which lets us use simple sine and cosine on the 40 degree bearing angle at P.

Bearing of Q from P=040∘,PQ=16 km\text{Bearing of } Q \text{ from } P = 040^\circ, \quad PQ = 16\ \text{km}
∠PRQ=90∘\angle PRQ = 90^\circ
2

Find the legs PR and QR with right-angle trig

In the right-angled triangle PQR the bearing angle of 40 degrees sits at P, measured from the north line. The east leg PR is opposite this angle, so it uses sine, and the north leg QR is adjacent, so it uses cosine. These give how far east R is from P and how far Q is above R.

PR=16sin⁡40∘≈10.28 kmPR = 16\sin 40^\circ \approx 10.28\ \text{km}
QR=16cos⁡40∘≈12.26 kmQR = 16\cos 40^\circ \approx 12.26\ \text{km}
3

Locate S using its bearing from R

From R draw a fresh north line and place S 12 km away on a bearing of 120 degrees. A bearing of 120 degrees points south of east, so S is both east of R and below it. Splitting the 12 km into east and south parts uses sine and cosine of 120 degrees measured from R's north line.

East of R=12sin⁡120∘≈10.39 km\text{East of } R = 12\sin 120^\circ \approx 10.39\ \text{km}
South of R=12∣cos⁡120∘∣=6 km\text{South of } R = 12\left|\cos 120^\circ\right| = 6\ \text{km}

R is already 10.28 km east of P, so S ends up about 20.67 km east of P and 6 km south of P.

4

Find the distance PS by Pythagoras

Relative to P, the point S is 20.67 km east and 6 km south. These two displacements are at right angles, so the straight-line distance PS is the hypotenuse of a right-angled triangle and comes straight from Pythagoras.

PS=20.672+62≈463.3≈21.5 kmPS = \sqrt{20.67^{2} + 6^{2}} \approx \sqrt{463.3} \approx 21.5\ \text{km}
5

Find the bearing of P from S

Standing at S, the point P lies to the north and west, so its bearing is in the last quarter, between 270 and 360 degrees. Work out the angle the line SP makes west of north, then subtract it from 360 degrees to read the bearing clockwise from north.

tan⁡θ=20.676⇒θ≈73.8∘\tan\theta = \frac{20.67}{6} \Rightarrow \theta \approx 73.8^\circ
Bearing of P from S=360∘−73.8∘≈286∘\text{Bearing of } P \text{ from } S = 360^\circ - 73.8^\circ \approx 286^\circ
6

Find the area of the ranch

The ranch PQRS is split along the diagonal PR into two triangles. Triangle PQR has the two perpendicular legs PR and QR, and triangle PRS uses the base PR with the 6 km that S sits below R as its height. Add the two half-base-times-height areas together.

Area=12(10.28)(12.26)+12(10.28)(6)\text{Area} = \tfrac{1}{2}(10.28)(12.26) + \tfrac{1}{2}(10.28)(6)
Area≈63.0+30.8≈93.9 km2\text{Area} \approx 63.0 + 30.8 \approx 93.9\ \text{km}^{2}

Final Result

PS is about 21.5 km, the bearing of P from S is about 286 degrees, and the area of the ranch PQRS is about 93.9 square kilometres.

Why this method works

Every bearings problem becomes easy once you turn each journey into east and north (or south) components using sine and cosine of the bearing angle. The right angle at R lets you split the 16 km trip into a 10.28 km eastward part and a 12.26 km northward part, and the 120 degree bearing splits the 12 km into 10.39 km east and 6 km south. Once all points are described by simple east and north distances from P, the straight-line distance PS is just Pythagoras, the bearing back to P is an inverse-tangent adjusted for the correct quarter of the compass, and the enclosed area is found by cutting the shape into right-angled triangles you can measure with half base times height.

✓

Coordinates from P: R at (10.28, 0), Q at (10.28, 12.26), S at (20.67, -6). Distance PS = sqrt(20.67^2 + 6^2) = 21.5 km, confirming the answer.