Solutions — Trig 6D Worksheet

Year 10 Mathematics Core · Cambridge Ch 6, §6D · Mr Wong

ANSWER KEY

Warm-up

Q1. Cardinal directions as true bearings:

a $000°$T
b $045°$T
c $090°$T
d $135°$T
e $180°$T
f $225°$T
g $270°$T
h $315°$T

Q2. Reverse bearings (±180°):

a $200°$T
b $082°$T
c $335°$T
d $164°$T

Q3. Bearings from $O$:

a $A=055°$T
b $B=135°$T
c $C=240°$T
d $D=340°$T

Part A — Reverse bearings

Q4. Bearing of $O$ from each point = forward bearing $\pm 180°$:

a $235°$T
b $315°$T
c $060°$T
d $160°$T

Part B — Bearings with trigonometry (2 d.p.)

5a 120°T from turning pt is $30°$ south of east ($60°$ east of south). East = $11\sin 60°=11\cos 30°\approx$ $9.53$ km east
5b South leg = $11\sin 30°=5.5$;  total south = $5+5.5=$ $10.5$ km south
6a 070°T is $70°$ east of north. East = $100\sin 70°\approx$ $93.97$ km east
6b North = $100\cos 70°\approx$ $34.20$ km north
7a 215°T is $35°$ west of south. West = $6\sin 35°\approx$ $3.44$ km west
7b South = $6\cos 35°\approx$ $4.91$ km south
8a 050°T leg: east = $5\sin 50°\approx$ $3.83$ km east
8b Total north = $4+5\cos 50°\approx$ $7.21$ km north
9. Leg 1: $E_1=4\sin 110°\approx 3.76$, $N_1=4\cos 110°\approx -1.37$. Leg 2: $E_2=3\sin 200°\approx -1.03$, $N_2=3\cos 200°\approx -2.82$. Totals: $E\approx 2.73$, $N\approx -4.19$. Bearing = $\tan^{-1}\!\left(\dfrac{2.73}{4.19}\right)$ in S–E quadrant $\Rightarrow$ $\approx 146.9°$T from start