Reverse bearings differ by \(180^\circ\), so the bearing of \(A\) from \(B\) is \(240^\circ\). Thus \(2x+50=240\), giving \(x=95\).

The two perpendicular legs are equal, so the angle between south and \(PR\) is \(45^\circ\). Measure clockwise from north: \(180^\circ+45^\circ=\boxed{225^\circ}\).

(a) AB measures about \(9.4\text{ cm}\) using the scale, so the actual distance is \(9.4\times2=\boxed{18.8\text{ km}}\).

(b) Measure clockwise from the north line at A to AB: the bearing is approximately \(\boxed{257^\circ}\).

(a) Use the 1 cm reference to preserve the drawing scale. \(PQ\approx9.0\) cm, so the distance is \(9(0.5)=\boxed{4.5\text{ km}}\).

(b) Measure clockwise from north at P to PQ: \(\boxed{110^\circ}\).

(a) Measure 80° clockwise from north and draw AB=6 cm. Mark its endpoint B. Use the 1 cm reference below the source diagram if printing or resizing.

(b) The reverse bearing differs by 180°: \(\boxed{310^\circ}\).

Add 180° to reverse the bearing: \(\boxed{287^\circ}\).

The reverse bearing from C to A is 104+180=284°. The triangle is equilateral, so each angle is 60°. CB lies 60° anticlockwise from CA in the diagram. Bearing \(=284-60=\boxed{224^\circ}\).

(a) West has bearing 270°; CH is 25° south of west, so \(\boxed{245^\circ}\). (b)(i) \(180-55-25=\boxed{100^\circ}\). (ii) Sine rule: \(BH=32\sin25^\circ/\sin100^\circ=13.73\ldots=\boxed{13.7\text{ km}}\). (c) The bearing BH from B is 90+100=190°. By cosine rule in ABH, \(\cos ABH=(11^2+BH^2-14^2)/(2\times11\times BH)\). Angle ABH≈67.9°, hence bearing BA≈190+67.9=\(\boxed{257.9^\circ}\). (d)(i) Speed=18.52 km/h; time=32/18.52 hours≈104 minutes. Arrival=\(\boxed{2{:}44\text{ pm}}\). (ii) The shortest distance to B occurs at the perpendicular foot on CH. Its distance from H is \(BH\cos55^\circ=\boxed{7.88\text{ km}}\) to 3 significant figures.

(a) KL measures 8.5 cm at the original scale, so \(8.5\times10=\boxed{85\text{ km}}\). (b) Measuring clockwise from north at K gives \(\boxed{065^\circ}\).

By the sine rule \(\sin\angle UWV=\frac{27.3\sin125^\circ}{62.4}\), giving \(\angle UWV=21.000679\ldots^\circ\).

The bearing of \(W\) from \(V\) is \(35^\circ\), so \(V\) from \(W\) is \(215^\circ\). Hence \(U\) from \(W\) is \(215+21.000679\ldots=236^\circ\).

(a) \(\angle CAB=90-38=52^\circ\). By the sine rule, \(\sin B=\frac{60\sin52^\circ}{87}\). The obtuse alternative for \(B\) would make the angle sum exceed 180°, so take the acute value.

\(C=180^\circ-52^\circ-\sin^{-1}\left(\frac{60\sin52^\circ}{87}\right)=95.08\ldots^\circ\approx95.1^\circ\).

(b) Keep the unrounded angle in \(AB=\frac{87\sin C}{\sin52^\circ}\approx110\text{ km}\). Total distance is about 257 km. Time \(=3+20/60=10/3\text{ h}\). Average speed \(=\frac{60+87+AB}{10/3}=77.1\text{ km/h}\).

Bearing of \(P\) from \(Q\) is \(90-39=51\) degrees. Reverse bearing \(51+180=231\), so 231°.

\(59+180=239\)°, written 239°.

Reverse bearings differ by 180°. Write \(x=2k,y=7k\). Then \(5k=180\) gives \(k=36\), hence \(y=252\)°.

Sine rule \(\sin B=17.6\sin25^\circ/12.8\). The acute value is 35.5°, so the required obtuse angle is \(B=144.5\)°. South to BC is \(180-112=68\)°. BA is therefore \(144.5-68=76.5\)° west of south. Reversing the bearing gives B from A: \(076.5\)° approximately (076° to the nearest degree).

Angle MLN = 60°, and N lies clockwise from M as seen from L. Bearing of N from L is \(103+60=163^\circ\). Reverse bearing \(163+180=343^\circ\).

(a) \(PQ\approx8.2\) cm on the original scale, giving \(8.2(4)=32.8\) km. (b) Clockwise from north: 065°. (c) Draw a ray 140° clockwise from north at P and mark X \(28/4=7\) cm from P, using the same scale.

\(105^\circ+180^\circ=285^\circ\).

\(274^\circ-180^\circ=94^\circ\), written as 094°.

(a) \(\theta=\tan^{-1}(350/450)=37.9^\circ\), so bearing037.9°.

(b) Take A as(0,0), east positive and north positive. \(B=(400\sin140^\circ,400\cos140^\circ)\). Thus \(D=(400\sin140^\circ+350,400\cos140^\circ+450)\). Pythagoras gives \(DA=\sqrt{(400\sin140^\circ+350)^2+(400\cos140^\circ+450)^2}=624\text{ m}\).

(c) \(t=\frac{400+350+450+DA}{3}=607.95\ldots\text{ s}\). Round to608 seconds =10 minutes8 seconds.

(a)(i) By the cosine rule, \(\cos\angle CBD=\frac{192^2+168^2-287.9^2}{2(192)(168)}\). Hence angle CBD=106.015…°≈106.0°. (ii) Bearing \(360+38-106.015\ldots=291.985\ldots^\circ\), or 292.0°.

(iii) Angle ABD=90°−38°=52°. By the sine rule, \(\sin\angle BAD=\frac{168\sin52^\circ}{205.8}\), so angle BAD=39.965…°. The obtuse alternative would make the triangle’s angle sum exceed 180°. Bearing D from A is \(270^\circ+39.965\ldots^\circ=310.0^\circ\).

(b)(i) \(A=\tfrac12(192)(168)\sin106.015\ldots^\circ=15502\ldots\text{ m}^2\approx15500\text{ m}^2\). (ii) Use the unrounded area: \(\text{cost}=\frac{A}{10000}(35750)\approx55420\) dollars, or \$55400 to the nearest \$100.

Reverse direction adds180°: \(128+180=308\)°.

\(102+180=282\)°.

(a) \(\angle LPM=201-127=74^\circ\). Sine rule \(LM=\frac{248\sin74^\circ}{\sin42^\circ}=356.2726\ldots\approx356\) km. (b)(i) Angle PLM=180−74−42=64°. Perpendicular distance \(LM\sin64^\circ\approx320\) km. (ii) Time \(248/40=6.2\) h=6 h 12 min. 20:45+6:12=02:57 next day.

Reverse the direction by 180°: 140+180=320°.

Reverse the direction:227−180=47°, written 047°.

(a)(i) Reverse bearing 110+180=290°. (ii) ABC=290−280=10°. Sine rule gives \(\sin C=50\sin10^\circ/38\). The diagram selects C≈13.2073°, so \(A=180-10-13.2073\ldots\approx156.8^\circ\).

(iii) The perpendicular distance is \(50\sin10^\circ=8.68241\ldots\approx8.68\) km. (b)(i) \(x(x-25)=2200\) expands to the required equation. (ii) \(x=\frac{25\pm\sqrt{625+8800}}2=\frac{25\pm\sqrt{9425}}2\), giving −36.04 and 61.04. The park length is 61.04 m.

(a) \(\frac12(70)(100)+\frac12(100)(110)\sin40^\circ\approx7035.3\) m², or 7040 m² to 3 significant figures.

(b) \(BC=\sqrt{70^2+100^2}=122.066\ldots\) m; \(AD=\sqrt{100^2+110^2-2(100)(110)\cos40^\circ}=72.436332\ldots\) m. Perimeter=\(70+110+BC+AD\approx374.5\) m. (c) Perpendicular height=\(100\sin40^\circ=64.2788\ldots\approx64.3\) m.

(d) \(\angle ABC=\tan^{-1}(100/70)=55.008\ldots^\circ\). C is west of south from B, so bearing=\(180+55.008\ldots\approx235^\circ\).

(a)(i) The bearing of A from C is \(135^\circ+180^\circ=315^\circ\). B is a further \(40^\circ\) anticlockwise, so its bearing from C is \(275^\circ\). (ii) The reverse bearing is \(275^\circ-180^\circ=095^\circ\). (b) \(AB=\sqrt{510^2+720^2-2(510)(720)\cos40^\circ}=464.668657\ldots\) km, giving 464.7 km. (c) The sine rule gives \(\sin B=\frac{510\sin40^\circ}{AB}\). Thus \(B\approx44.9^\circ\). The acute value matches the diagram and side lengths.

(a) \(BC=\sqrt{220^2-180^2}=126.4911\ldots\approx126\) m. (b) \(CD=\sqrt{180^2+170^2-2(180)(170)\cos33^\circ}\approx99.9\) m. (c) The perpendicular distance is \(170\sin33^\circ\approx92.6\) m. (d) \(\angle CAB=\cos^{-1}(180/220)\). Bearing \(=47^\circ+33^\circ+\angle CAB\approx115.1^\circ\). (e) \(A=\frac12(180)\sqrt{220^2-180^2}+\frac12(180)(170)\sin33^\circ\approx19700\) m².

(a) \(\frac{AB}{\sin85^\circ}=\frac{240}{\sin50^\circ}\Rightarrow AB=\frac{240\sin85^\circ}{\sin50^\circ}\approx312\) m. (b) \(12000=\frac12(180)(240)\sin\angle CAD\), so \(\angle CAD=\sin^{-1}(5/9)=33.74898\ldots^\circ\approx33.75^\circ\). The acute value agrees with the diagram. (c) \(\angle CAB=180^\circ-50^\circ-85^\circ=45^\circ\). Hence \(\angle DAB=33.74898\ldots^\circ+45^\circ\). The cosine rule gives \(BD=\sqrt{180^2+AB^2-2(180)(AB)\cos\angle DAB}\approx328\) m. (d)(i) \(030^\circ+33.75^\circ+45^\circ=108.75^\circ\). (ii) \(108.75^\circ+180^\circ=288.75^\circ\).

(a) \(\cos\angle ABL=\frac{40^2+61.1^2-92.1^2}{2(40)(61.1)}\), so \(\angle ABL=130.1111\ldots^\circ\approx130.1^\circ\). (b) The bearing of BA is \(90^\circ+\angle ABL\), so the bearing of AB is \(\angle ABL-90^\circ\). By the sine rule, \(\angle BAL=\sin^{-1}\left(\frac{40\sin\angle ABL}{92.1}\right)\approx19.39^\circ\). Add this to the bearing of AB: the bearing of AL is approximately 059.5°. (c) \(\angle LBC=180^\circ-\angle ABL\). The chord \(BC=2(40)\cos\angle LBC\approx51.54\) km. Time \(BC/28\approx1.841\) hours, approximately 110 minutes, or 1 h 50 min.

Angle CBD = 120° − 90° = 30°. By the sine rule, \(\sin D/6=\sin30^\circ/8\Rightarrow D=22.024\ldots^\circ\). The obtuse alternative cannot fit the triangle. Thus \(C=180^\circ-30^\circ-D=127.975\ldots^\circ\). CD is below CB, so bearing \(270^\circ-C=142.024\ldots^\circ\approx142^\circ\).

(a) LPQ = 142° − 110° = 32°. \(LQ=\sqrt{58^2+74^2-2(58)(74)\cos32^\circ}=39.501\ldots\approx39.5\) km. (b) \(\sin Q=58\sin32^\circ/LQ\Rightarrow Q=51.085\ldots^\circ\approx51.1^\circ\). (c)(i) 322°. (ii) \(322^\circ+51.085\ldots^\circ-360^\circ=13.085\ldots^\circ\), giving 013.1°. (d) \(74/(2.25\times1.85)=17.777\ldots\approx17.8\) knots. (e) Perpendicular distance \(58\sin32^\circ=30.735\ldots\approx30.7\) km.

(a) \(\tfrac12(8)(5)\sin150^\circ=10\) km². (b) BC is 30° east of north. Its reverse bearing is \(30+180=210^\circ\).

(a) \(ML=\sqrt{4^2+9^2-2(4)(9)\cos108^\circ}=10.9201\ldots\approx10.9\) km. (b)(i) \(-9\cos125^\circ=5.16219\ldots\approx5.16\) km south. (ii) Bearing K to M = \(125+108=233^\circ\). Reverse bearing \(233-180=053^\circ\).

(a) \(\sin\angle CAD=10\sin95^\circ/12\Rightarrow CAD=56.1154\ldots\approx56.1^\circ\). (b) \(BC=\sqrt{12^2+17^2-2(12)(17)\cos30^\circ}=8.92534\ldots\approx8.93\) km. (c)(i) \(40+56.1154\ldots+30=126.1154\ldots\approx126^\circ\). (ii) 306°. (d) Choose the obtuse solution: \(ACB=180^\circ-\sin^{-1}[17\sin30^\circ/BC]=107.7595\ldots^\circ\). \(DCA=180-95-56.1154\ldots=28.8846\ldots^\circ\). Therefore \(BCD=ACB+DCA=136.6441\ldots\approx137^\circ\).

(a)(i) AC is about 7.3 cm, so \(7.3(20)=146\) km. (ii) Bearing approximately 060°. (iii) Draw 7 cm from C on bearing 150°. (iv) Reverse bearing \(150+180=330^\circ\). (v) Area scale is squared: \(0.15(20)^2=60\) km². (b) Flight time = 2 h 15 min = 2.25 h: \(1500/2.25=666.666\ldots\approx667\) km/h. (c) \(\cos Q=(1125^2+790^2-1450^2)/(2\times1125\times790)\), giving \(Q\approx96.9^\circ\).

(a) \(\cos C=(2.7^2+4.5^2-5^2)/(2\cdot2.7\cdot4.5)\), giving \(C\approx84.0^\circ\). (b) From the diagram, B is 84° anticlockwise from CA: \(220-84=136^\circ\), so 136°.

(a) Sine rule: \(\sin R/50=\sin140^\circ/100\Rightarrow R=\sin^{-1}(\tfrac12\sin140^\circ)\approx18.7^\circ\). The obtuse alternative is impossible with Q already 140°. (b) Reverse bearing R to Q = 280°. From the diagram, RP is 18.7° anticlockwise from RQ: \(280-18.7=261.3^\circ\), approximately 261°.

(a) Cosine rule: \(\cos A=(180^2+115^2-90^2)/(2(180)(115))\), giving \(A\approx24.9874^\circ\approx25.0^\circ\). (b)(i) \(150^\circ-A\approx125^\circ\). (ii) Add 180°: 305°. (c) East displacement \(180\sin(150^\circ-A)\approx147\) km. (d) Sine rule: \(TR=90\sin30^\circ/\sin70^\circ\approx47.9\) km. (e) \(90\text{ km}=9\,000\,000\text{ cm},\ n=9\,000\,000/4.5=2\,000\,000\).