(a) Area \(=\frac12(450)(600)\sin57^\circ=113220.526\ldots\text{ m}^2\approx113000\text{ m}^2\).

(b) The sine rule in \(ACD\) gives \(\sin\angle ADC=\frac{450\sin34^\circ}{290}\). The acute angle shown at \(D\) is \(60.19397\ldots^\circ\). Thus \(\angle CAD=180-34-60.19397\ldots=85.80603\ldots^\circ\approx85.8^\circ\).

(c) \(\angle BAD=57+85.80603\ldots=142.80603\ldots^\circ\). By the cosine rule, \(BD=\sqrt{600^2+290^2-2(600)(290)\cos142.80603\ldots^\circ}=849.302\ldots\) m. The height difference is \(250-45=205\) m. Therefore \(\theta=\tan^{-1}(205/849.302\ldots)=13.5702\ldots^\circ\approx13.6^\circ\).

By the sine rule, \(\sin y/8.1=\sin105^\circ/16.8\). Thus \(y=\sin^{-1}(8.1\sin105^\circ/16.8)=27.756456\ldots^\circ\), giving \(\boxed{y=27.8^\circ}\). The supplementary angle is impossible because the triangle already has an obtuse angle.

(a) \(A=\frac12(60.4)(52.3)\sin26^\circ=692.389691\ldots\), giving \(\boxed{692\text{ mm}^2}\).

(b) \(\cos\angle ACD=\frac{60.4^2+31.5^2-65.1^2}{2(60.4)(31.5)}\). Thus \(\angle ACD=83.929619\ldots^\circ\), giving \(\boxed{83.9^\circ}\).

(a) Bearings are clockwise from north: \(360^\circ-103^\circ=\boxed{257^\circ}\).

(b) By the cosine rule in ADC, \(CD^2=38.2^2+86.8^2-2(38.2)(86.8)\cos79^\circ\). Thus \(CD=87.909762\ldots\), giving \(\boxed{87.9\text{ m}}\).

(c) In ABC, the sine rule gives \(\sin C/50.3=\sin112^\circ/86.8\). Hence \(C=\sin^{-1}(50.3\sin112^\circ/86.8)=32.499797\ldots^\circ\).

Then \(\angle BAC=180^\circ-112^\circ-C=35.500202\ldots^\circ\). Drop the perpendicular from B to AC. Its length is \(d=50.3\sin(\angle BAC)=29.209503\ldots\), so the shortest distance is \(\boxed{29.2\text{ m}}\).

(a) By the sine rule, \(\sin A=\frac{15\sin120^\circ}{15\sqrt3}=\frac12\). The angle A cannot be \(150^\circ\), since B is already \(120^\circ\); hence \(A=30^\circ\). Then \(C=180^\circ-120^\circ-30^\circ=30^\circ\). Two angles are equal, so the opposite sides AB and BC are equal: ABC is isosceles.

(b) Drop the perpendicular from B to AC. Its length is \(15\sin30^\circ=\boxed{7.5\text{ cm}}\).

The largest angle is opposite the 11 cm side. By the cosine rule, \(\cos\theta=\frac{5^2+7^2-11^2}{2(5)(7)}=-47/70\). Thus \(\theta=132.177418\ldots^\circ\), giving \(\boxed{132.2^\circ}\).

(a) The bearing of A from C is \(145^\circ+180^\circ=325^\circ\). Moving anticlockwise by \(38^\circ\) gives \(\boxed{287^\circ}\).

(b) \(AB=\sqrt{65^2+95^2-2(65)(95)\cos38^\circ}=59.313296\ldots\), giving \(\boxed{59.3\text{ m}}\).

(c) By the sine rule, \(\sin A=95\sin38^\circ/AB\). Since A is obtuse, take \(A=180^\circ-\sin^{-1}(95\sin38^\circ/59.313296\ldots)=99.570039\ldots^\circ\). The bearing is \(145^\circ+A=\boxed{244.6^\circ}\).

(a) Cosine rule in ABD: \(AD^2=9.6^2+12.1^2-2(9.6)(12.1)\cos56^\circ\). Thus \(AD=10.423929\ldots\), giving \(\boxed{10.4\text{ cm}}\).

(b) Sine rule in BCD gives \(\sin D/15.8=\sin47^\circ/12.1\). The acute inverse-sine value is \(72.744366\ldots^\circ\); choose the obtuse solution: \(D=180^\circ-72.744366\ldots^\circ=\boxed{107.3^\circ}\).

(c) Retain the unrounded angle: \(\angle CBD=180^\circ-47^\circ-107.255633\ldots^\circ=25.744366\ldots^\circ\). Area \(=\frac12(9.6)(12.1)\sin56^\circ+\frac12(12.1)(15.8)\sin25.744366\ldots^\circ=89.670659\ldots\), giving \(\boxed{89.7\text{ cm}^2}\).

(a) \(AB=12\tan34^\circ=8.094102\ldots\), so \(\boxed{8.09\text{ m}}\).

(b) Add the two triangle areas: \(\frac12(10)(12)\sin56^\circ+\frac12(12)(12\tan34^\circ)=98.306867\ldots\), giving \(\boxed{98.3\text{ m}^2}\).

(c) Cosine rule gives \(BC=\sqrt{10^2+12^2-2(10)(12)\cos56^\circ}=10.478249\ldots\). Also \(AD=12/\cos34^\circ=14.474615\ldots\). Perimeter \(=AB+BC+CD+DA=43.046966\ldots\), giving \(\boxed{43.0\text{ m}}\).

(d) Drop the perpendicular from B to AD. Its length is \(12\sin34^\circ=6.710314\ldots\), giving \(\boxed{6.71\text{ m}}\).

\(A=\frac12ab\sin C=\frac12(6)(10)\sin30^\circ=30(1/2)=\boxed{15\text{ cm}^2}\).

\(A=\frac12(8)(9)\sin50^\circ=27.577599\ldots\), giving \(\boxed{27.6\text{ cm}^2}\).

(a) In ABC, cosine rule gives \(\cos y=(10^2+14^2-13^2)/(2\times10\times14)=127/280\). Hence \(\boxed{y=63.0^\circ}\).

(b) In ADC, \(\angle CAD=180^\circ-97^\circ-38^\circ=45^\circ\). Sine rule gives \(AD=14\sin38^\circ/\sin97^\circ=8.683989\ldots\text{ cm}\). Thus \(\angle BAD=45^\circ+y\). Using unrounded values, \(BD=\sqrt{10^2+AD^2-2(10)(AD)\cos(45^\circ+y)}=15.138010\ldots\), giving \(\boxed{15.1\text{ cm}}\).

(a) Cosine rule: \(BC=\sqrt{300^2+112^2-2(300)(112)\cos140^\circ}=392.456604\ldots\), giving \(\boxed{392\text{ m}}\).

(b) Sine rule: \(\sin\angle DBC=112\sin140^\circ/BC\). Using unrounded BC, \(\boxed{\angle DBC=10.6^\circ}\).

(c) Area BCD \(=\frac12(300)(112)\sin140^\circ\). Subtract this from 35900 to find area ABD. The shortest distance h is perpendicular to AB, so \(\frac12(180)h=35900-\frac12(300)(112)\sin140^\circ\). Hence \(h=278.901868\ldots\), giving \(\boxed{279\text{ m}}\).

(a) Sine rule in the upper triangle: \(\sin x/41.3=\sin102^\circ/64.5\), giving \(\boxed{x=38.8^\circ}\). The other inverse-sine angle is impossible with the 102° angle.

(b) Cosine rule in the lower triangle: \(\cos y=(64.5^2+52.1^2-70.2^2)/(2(64.5)(52.1))\), giving \(\boxed{y=73.2^\circ}\).

\(A=\frac12(4.9)(5.6)\sin23^\circ=5.360831\ldots\), giving \(\boxed{5.36\text{ cm}^2}\).

(a) Sine rule: \(AC=17.2\sin54^\circ/\sin68^\circ=\boxed{15.0\text{ cm}}\) to 3 significant figures.

(b) Cosine rule in AMC: \(AM=\sqrt{AC^2+12.8^2-2(AC)(12.8)\cos68^\circ}\). Using unrounded AC gives \(\boxed{15.7\text{ cm}}\). (c) Drop a perpendicular from A to BC. Height \(=17.2\sin54^\circ=\boxed{13.9\text{ cm}}\).

\(A=\frac12(6.7)(5.9)\sin81^\circ=\boxed{19.5\text{ cm}^2}\).

(a) Cosine rule: \(BD=\sqrt{10.4^2+6.5^2-2(10.4)(6.5)\cos64^\circ}=9.546\ldots=\boxed{9.55\text{ cm}}\). (b)(i) 180−26−42=\(\boxed{112^\circ}\). (ii) Sine rule: \(CD=BD\sin42^\circ/\sin112^\circ=\boxed{6.89\text{ cm}}\). (c) The perpendicular height=\(AD\sin64^\circ=6.5\sin64^\circ=\boxed{5.84\text{ cm}}\).

(a) Pythagoras: \(GH=\sqrt{4^2+1.5^2}=\boxed{4.27\text{ m}}\). (b) Angle west of south=\(\tan^{-1}(50/120)=22.6^\circ\). Bearing=\(\boxed{180+22.6=202.6^\circ}\). (c) ADB=180−25−15=140°. Sine rule gives \(AB=5\sin140^\circ/\sin25^\circ\). Since BC=AB, perimeter=\(2AB+10=\boxed{25.2\text{ cm}}\). (d) In PSR, \(PR^2=11^2+8^2-2(11)(8)\cos110^\circ\). In PQR, \(\cos PQR=(10^2+14^2-PR^2)/(2(10)(14))\). Hence \(\boxed{PQR=79.5^\circ}\).

(a) Cosine rule: \(AC^2=420^2+830^2-2(420)(830)\cos106^\circ\). Thus \(AC=1028.335\ldots\text{ m}\), rounding to 1028 m. (b) \(\sin C=1150\sin62^\circ/AC\). The acute solution is about 80.9°; take the obtuse value \(\boxed{C=99.1^\circ}\). (c) Area=\(\frac12(420)(830)\sin106^\circ=167547.9\ldots\text{ m}^2\). Price per hectare \(=\frac{41500\times10000}{\frac12(420)(830)\sin106^\circ}=2476.90\ldots\), so \(\boxed{\$2477}\).

(a)(i) \(CBD=180-60-39=81^\circ\). (ii) Sine rule: \(CD=129\sin81^\circ/\sin60^\circ=\boxed{147\text{ m}}\). (iii) \(\cos ABD=(85^2+129^2-72^2)/(2\times85\times129)\), so ABD=31.5819…°, rounding to 31.6°. (iv) Perpendicular height=\(85\sin ABD=\boxed{44.5\text{ m}}\). (v) Bearing of C from B=180−ABD−81=67.418…°. Reverse bearing=\(\boxed{247.4^\circ}\) to 1 decimal place.

(vi) \(A_{ABD}=\frac12(85)(129)\sin ABD\), \(A_{BCD}=\frac12(129)(CD)\sin39^\circ\). Total area=8843.15… m², so trees=\((A_{ABD}+A_{BCD})\times1100/10000=972.746\ldots\approx\boxed{973}\). (b) Use maximum area 9450 and minimum length 75: \(9450/75=\boxed{126\text{ cm}}\).

(a) Cosine rule: \(AC=\sqrt{8.7^2+11.4^2-2(8.7)(11.4)\cos119^\circ}=17.37287\ldots\text{ cm}\), rounding to 17.37. (b) Sine rule in ACE: \(\sin AEC=AC\sin20^\circ/10.9\). The diagram requires the obtuse angle, so \(AEC=180^\circ-\sin^{-1}(AC\sin20^\circ/10.9)\). Then \(ACE=180^\circ-20^\circ-AEC=\boxed{13.0^\circ}\). (c) In right triangleADC, \(AD=AC\sin ACE\), \(CD=AC\cos ACE\). Thus perimeter=\(8.7+11.4+AD+CD=\boxed{40.9\text{ cm}}\), using unrounded AC and ACE.

(a) \(AB=12.3\cos31^\circ=10.5431578\ldots=10.54\text{ cm}\).

(b) \(\angle DAC=\cos^{-1}(12.3/16.5)=41.8018442\ldots^\circ=41.80^\circ\).

(c) Use unrounded values: \(BC=12.3\sin31^\circ=6.3349683\), \(CD=\sqrt{16.5^2-12.3^2}=10.9981817\). Angle \(BCD=149^\circ\).

\(BD=\sqrt{BC^2+CD^2-2(BC)(CD)\cos149^\circ}=16.7491757\ldots=16.7\text{ cm}\).

(d) \(\cos\angle CBD=\frac{BD^2+BC^2-CD^2}{2(BD)(BC)}\). Thus \(\angle CBD=19.7669231\ldots^\circ=19.8^\circ\).

(e) The shortest distance is perpendicular to \(BD\): \(BC\sin\angle CBD=2.1424527\ldots=2.14\text{ cm}\).

\(2143=\frac12(92.5)(71)\sin x\). Thus \(x=\sin^{-1}\left(\frac{2(2143)}{92.5(71)}\right)=40.7^\circ\).

(a)(i) Cosine rule: \(AB=\sqrt{4.8^2+5.6^2-2(4.8)(5.6)\cos20.4^\circ}=2.0029279\ldots=2.00\text{ m}\).

(ii) \(AX=AB\tan64^\circ=4.1066108\ldots=4.11\text{ m}\), using unrounded \(AB\). (iii) \(\frac12(4.8)(2.9)=6.96\text{ m}^2\).

(b) \(\angle RMP=180-75=105^\circ\), so \(\angle RPM=180-105-30=45^\circ\). Sine rule \(PM=\frac{8\sin30^\circ}{\sin45^\circ}=4\sqrt2\). Thus \(PQ=2PM=8\sqrt2=11.3\text{ cm}\).

Sine rule \(\frac{\sin y}{8.3}=\frac{\sin105^\circ}{16.2}\). Thus \(y=\sin^{-1}\left(\frac{8.3\sin105^\circ}{16.2}\right)=29.7^\circ\). The obtuse alternative is impossible alongside the 105° angle.

Sine rule \(\sin R=\frac{18\sin42^\circ}{13.5}\). Take the obtuse solution \(R=180^\circ-\sin^{-1}\left(\frac{18\sin42^\circ}{13.5}\right)=116.9^\circ\).

(a) \(AB=83.2/\tan38^\circ=106.4911\ldots\text{ m}\). \(AC=AB+54.5\). \(\angle ACD=\tan^{-1}(83.2/AC)=27.3^\circ\).

(b) Mark the midpoint of hypotenuse \(FG\). A right angle is the angle in a semicircle, so the hypotenuse is a diameter.

(c) Largest \(LMN\) angle \(=180(6/15)=72^\circ\). Largest \(PQR\) angle is opposite side 6: \(R=\cos^{-1}\left(\frac{4^2+5^2-6^2}{2(4)(5)}\right)=82.81924\ldots^\circ\). Difference \(=82.81924\ldots-72=10.8^\circ\).

(a)(i) \(\tan^{-1}(2.8/7.1)=21.5^\circ\). (ii) \(PR=2.8/\tan21^\circ\). Then \(PQ=\sqrt{PR^2+7.1^2}=10.2\text{ m}\).

(b) Sine rule \(\sin M=\frac{16.7\sin32^\circ}{9.1}\). Acute \(M=76.5^\circ\).

(c)(i) \(\frac12(12.3)(21.5)\sin A=62.89\). Thus \(A=\sin^{-1}\left(\frac{2(62.89)}{12.3(21.5)}\right)\approx28.4^\circ\).

(ii) Use the unrounded angle: \(BC=\sqrt{12.3^2+21.5^2-2(12.3)(21.5)\cos A}=12.2\text{ cm}\). (iii) \(BD=AD-AB=21.5\cos A-12.3=6.61\text{ cm}\).

(a)(i) Sine rule \(\sin B=\frac{29.5\sin51.6^\circ}{35.3}\) gives \(B=40.9^\circ\). The obtuse alternative cannot form the given triangle. (ii) \(\frac12(45)(29.5)\sin51.6^\circ=520\text{ cm}^2\).

(b)(i) Bisect the isosceles triangle: \(SQ=64\sin28^\circ\). Cosine rule \(SR=\sqrt{SQ^2+47^2-2(SQ)(47)\cos60^\circ}=41.2\text{ cm}\). (ii) Altitude \(32\cos28^\circ=28.3\text{ cm}\).

(a) Cosine rule \(\cos\angle ACD=\frac{14^2+12^2-9^2}{2(14)(12)}\) gives \(\angle ACD=39.6^\circ\). (b) Sine rule \(BC=\frac{14\sin25^\circ}{\sin123^\circ}=7.054803\ldots=7.05\text{ km}\).

(c) \(BC\sin32^\circ=3.74\text{ km}\). (d) Using unrounded values, \(\angle BCD=32^\circ+\angle ACD\). \(BD=\sqrt{12^2+BC^2-2(12)(BC)\cos\angle BCD}=11.8\text{ km}\). (e) \(270^\circ+\angle ACD=309.6^\circ\).

(a) Cosine rule: \(CB=\sqrt{6.4^2+10.9^2-2(6.4)(10.9)\cos38^\circ}=7.06\text{ cm}\). (b)(i) \(\angle ABD=38^\circ\) (alternate angles), so \(\angle ADB=180-45-38=97^\circ\). (ii) \(AB=10.9\sin97^\circ/\sin45^\circ=15.3\text{ cm}\).

(c) Height \(h=10.9\sin38^\circ\). With unrounded AB, \(A=\frac12(6.4+AB)h=72.8\text{ cm}^2\).

(a) \(BC=\sqrt{82^2+55^2-2(82)(55)\cos76^\circ}=87.0\text{ m}\). (b) \(\sin C=82\sin76^\circ/BC\) gives \(C=66.2\)°; the supplementary angle is impossible since it plus 76° exceeds 180°.

(c) CG is perpendicular to AB. \(AG=55\cos76^\circ=13.3\text{ m}\). (d) \(\frac12(90)(60)\sin Q=\frac12(82)(55)\sin76^\circ\). Therefore \(Q=54.1\)° or \(125.9\)°.

\(C=180-115-35=30\)°. By the sine rule, \(BC=7\sin115^\circ/\sin30^\circ=12.7\text{ cm}\).

(a) \(\cos C=\frac{15^2+8^2-20^2}{2(15)(8)}=-0.4625\) gives \(C=117.5485\ldots\)°, hence 117.5°. (b) \(\frac12(15)(8)\sin C=53.2\text{ m}^2\). (c) \(AQ=\sqrt{15^2+4^2}=15.5\text{ m}\).

(d) Height difference 1 m and horizontal separation 8 m give \(\theta=\tan^{-1}(1/8)=7.1^\circ\). (e) Midpoint height 3.5 m. Its ground projection is the midpoint N of BC. Median \(AN=\sqrt{\frac{2(15^2)+2(20^2)-8^2}4}=\sqrt{296.5}\). Thus \(\tan^{-1}(3.5/AN)=11.5^\circ\).

(a)(i) Cosine rule: \(\cos P=\frac{39.4^2+46.5^2-38.2^2}{2(39.4)(46.5)}\) gives \(P=52.0^\circ\). (ii) Perpendicular height \(h=46.5\sin P=36.6\) cm. (b)(i) \(AG=\sqrt{29^2+21^2+20^2}=41.0\text{ cm}\). (ii) \(\theta=\sin^{-1}(20/AG)=29.2^\circ\).

(c) Bearing of M from K is 190°, so angle MKL = \(190-155=35^\circ\). From M, the bearing of L is \(10+96=106^\circ\), so from L the bearing of M is \(106+180=286^\circ\). Sine rule: \(LM=\frac{112\sin35^\circ}{\sin96^\circ}=64.6\text{ km}\).

(a) \(\cos\angle ADB=\frac{8.6^2+9.7^2-4.4^2}{2(8.6)(9.7)}\) gives 27.0°. (b) In triangle BCD, \(\angle BCD=180-65-42=73^\circ\). Sine rule gives \(DC=\frac{9.7\sin65^\circ}{\sin73^\circ}=9.19\text{ cm}\). (c) Perpendicular distance \(h=DC\sin42^\circ=6.15\) cm, using unrounded DC.

The largest angle is opposite the longest side, 11 cm. Cosine rule: \(\cos A=(10^2+9^2-11^2)/(2(10)(9))=1/3\). \(A=\cos^{-1}(1/3)=70.5^\circ\).

(a) Angle AXC = 102°. Cosine rule \(AC=\sqrt{10.6^2+6.4^2-2(10.6)(6.4)\cos102^\circ}=13.5\) cm. (b) Angle BAX = 44°. Sine rule \(BX=10.6\sin44^\circ/\sin58^\circ=8.68\) cm. (c) Total area \(\tfrac12(10.6)(BX+6.4)\sin78^\circ=78.2\) cm², using the unrounded value of BX.

(a) Cosine rule \(BD=\sqrt{16^2+19^2-2(16)(19)\cos57^\circ}=16.907379\ldots=16.9\) m. (b) Sine rule \(\sin C=BD\sin75^\circ/32\) gives C ≈ 30.68° (the obtuse alternative is impossible with angle D = 75°). Therefore \(\angle CBD=180-75-C=74.3^\circ\). (c) Using unrounded values, \(A=\tfrac12(16)(19)\sin57^\circ+\tfrac12(BD)(32)\sin(\angle CBD)=388\) m². (d) Perpendicular distance \(16\sin57^\circ=13.4\) m.

(a) In right triangle ABD: \(AB=14\cos35^\circ\), \(AD=14\sin35^\circ\). In BCD, angle C = 75°. Sine rule \(BC=14\sin60^\circ/\sin75^\circ\) and \(CD=14\sin45^\circ/\sin75^\circ\). Sum \(AB+BC+CD+AD=42.3\) cm. (b)(i) \(3l^2=8.5^2\), so \(l=8.5/\sqrt3=4.91\) cm. (ii) \(\sin\theta=l/8.5=1/\sqrt3\), hence \(\theta=35.3^\circ\).

Land area \(\frac12(800)(2300)\sin30^\circ=460000\) m². \(460000/400=1150\) houses.

(a)(i) In right triangle ADE, \(AD=46.1\tan64^\circ=94.519\ldots\) m, giving 94.5 m. (ii) Sine rule: \(\sin\angle BAC=56.5\sin94^\circ/78.4\), so \(\angle BAC=46.0^\circ\). (iii) AD is the longest side, so the largest angle is ACD. Cosine rule \(\cos\angle ACD=\frac{38.6^2+78.4^2-AD^2}{2(38.6)(78.4)}\) gives \(\angle ACD=102.4^\circ\) using unrounded AD (102.3° using 94.5 m is also accepted). (b) \(\frac12(34.3)d=\frac12(21.5)(27.6)\sin111^\circ\). Hence \(d=16.2\) cm.

(a) Sine rule in ADC: \(AD=\frac{12\sin50^\circ}{\sin100^\circ}=9.33434\ldots\approx9.33\) cm. (b) Cosine rule in ABC: \(\cos\angle BAC=\frac{11^2+12^2-8^2}{2(11)(12)}=67/88\). Hence \(\angle BAC=40.415439\ldots^\circ\approx40.42^\circ\). (c) DAC = 30°. Sum the triangle areas: \(\frac12(12)(AD)\sin30^\circ+\frac12(12)(11)\sin\angle BAC=70.79248\ldots\approx70.8\) cm². (d) Drop a perpendicular from B to AC: \(d=11\sin\angle BAC=7.13158\ldots\approx7.13\) cm.

(a)(i) Sine rule: \(QR=\frac{6\sin25^\circ}{\sin72^\circ}=2.67\text{ cm}\). (ii) Cosine rule: \(PS=\sqrt{6^2+7.4^2-2(6)(7.4)\cos34^\circ}=4.14\text{ cm}\). (iii) Angle SQR = 83°. \(A=\frac12(6)(QR)\sin83^\circ+\frac12(6)(7.4)\sin34^\circ=20.4\text{ cm}^2\). (b)(i) \(AG=\sqrt{20^2+18^2+16^2}=\sqrt{980}=31.30495\ldots\) cm. Thus \(GX=40-\sqrt{980}=8.70\text{ cm}\). (ii) \(\sin\theta=16/\sqrt{980}\), so \(\theta=30.7^\circ\).

(a) Cosine rule: \(BC^2=80^2+115^2-2(80)(115)\cos72^\circ\), so BC = 118.06… m, hence 118.1 m.

(b) Using BC =118.1 m from(a), the sine rule gives \(\sin B=115\sin72^\circ/118.1\). Thus B =67.8336…°, or67.8°. This is the acute angle because side AC is shorter than side BC.

(c)(i) Bearing AB is \(147^\circ-72^\circ=75^\circ\), so bearing BA = 255°. (ii) Bearing CA = 327°. Angle ACB = 180° − 72° − B. Bearing CB = 327° + angle ACB − 360° = 7.2° to 1 decimal place, written 007.2°.

(d) \(\frac{115}{35}(3.6)=11.8\text{ km/h}\). (e) The perpendicular distance is \(80\sin72^\circ=76.1\text{ m}\).

(a) \(QR^2=150^2+120^2-2(150)(120)\cos25^\circ\), so QR = 65.4 m.

(b) Angle PSR = 180° − 55° − 45° = 80°. By the sine rule, \(RS=\frac{150\sin55^\circ}{\sin80^\circ}=125\text{ m}\) (3 significant figures).

(c) \(A=\frac12(150)(120)\sin25^\circ+\frac12(150)(RS)\sin45^\circ=10400\text{ m}^2\) to3 significant figures, using unrounded RS.

(a)(i) Cosine rule in CGB: \(\cos x=\frac{5.3^2+11^2-6.9^2}{2(5.3)(11)}\), so x=29.5°. (ii) Angle C=54°+x, so angle B=180°−42°−54°−x. Sine rule gives \(AC=\frac{11\sin B}{\sin42^\circ}=13.4\text{ cm}\).

(b) Cross-sectional area \(15(2.5)=37.5\) cm²; volume per second \(37.5(20)=750\) cm³. In3600 seconds: \(750(3600)=2700000\) cm³ =2700 litres.

(a) By the sine rule, \(\frac{\sin x}{8}=\frac{\sin100^\circ}{9}\). Hence x=61.1°. The obtuse alternative is impossible alongside100°. (b) Third angle \(180^\circ-100^\circ-x=18.9^\circ\). Area \(\frac12(8)(9)\sin(180^\circ-100^\circ-x)=11.7\text{ cm}^2\) using unrounded x.

(a) In PSR, \(\frac{SR}{\sin53^\circ}=\frac{7.4}{\sin97^\circ}\), hence \(SR=7.4\sin53^\circ/\sin97^\circ=5.95\) cm.

(b) In PRQ, cosine rule: \(RQ=\sqrt{7.4^2+8.5^2-2(7.4)(8.5)\cos26^\circ}=3.73\text{ cm}\).

(a) \(ED=\sqrt{120^2+50^2}=130,\quad CD=\sqrt{170^2-150^2}=80\). Perimeter=100+170+80+130+50=530 m.

(b) Cosine rule: \(\cos B=\frac{100^2+150^2-120^2}{2(100)(150)}\), so angle ABD=52.9°. (c)(i) \(\cos\angle CBD=150/170\), giving 28.1°. (ii) D is west of north, so bearing=360°−28.072…°=331.9°.

(d) \(A=\frac12(50)(120)+\frac12(100)(150)\sin52.890995\ldots^\circ+\frac12(150)(80)=14981.168\ldots\) m². Divide by10000: 1.50 hectares.

(a) \((5-2)180/5=108\)°. (b)(i) Angle BOC=72°, so BOM=36° and \(BM=12\sin36^\circ=7.05\) cm. (ii)(a) OM⊥BC. Exterior angle MBX=72°, giving \(BX=BM/\cos72^\circ=22.8\) cm. (b) AB=2BM and AX=AB+BX≈36.932… cm. The perpendicular height from O to AB is the apothem \(12\cos36^\circ\). Hence \(A=\frac12(2(12\sin36^\circ)+BX)(12\cos36^\circ)=179\) cm².

(a)(i) \(AB=\sqrt{16.5^2+12.4^2-2(16.5)(12.4)\cos64^\circ}=15.7\) cm. (ii) Angle BCD=180°−53°−95°=32°; \(BC=\frac{12.4\sin53^\circ}{\sin32^\circ}=18.7\) cm. (b)(i) \(y=\frac{7.7}{3.8}\frac{180}{\pi}=116.1^\circ\). (ii) \(A=.5(3.8)(7.7)=14.63\approx14.6\) cm².

(a)(i) In ACD, angle ADC=180°−72°=108° and ACD=46°. Sine rule \(AC=\frac{58\sin108^\circ}{\sin46^\circ}=76.6832\ldots\approx76.7\) m. (ii) \(BC=AC\sin26^\circ,\quad BD=BC/\tan72^\circ=10.9\) m. (b)(i) Let EF=x, FG=2x. \(70=\frac12x(2x)\sin40^\circ\Rightarrow x=\sqrt{70/\sin40^\circ}=10.4\) m. (ii) Same sides and area give the same sine, so the different angle is180°−40°=140°.

\(A=\frac12(11)(13)\sin39^\circ=45.0\) cm².

Sine rule \(\sin x=\frac{12\sin39^\circ}8\). Acute inverse-sine value 70.73…°, so obtuse \(x=180-70.73\ldots=109.3\)°.

(a) Total time \((120+180+150)/3.6=125\) s. Times AB=30 s, BC=60 s, so CA takes 35 s. Speed \(150/35=4.29\) m/s.

(b) \(\cos A=\frac{120^2+150^2-180^2}{2(120)(150)}=.125\Rightarrow A=82.81924\ldots^\circ\approx82.8^\circ\). (c)(i) \(210-82.81924=127.2\)°. (ii) Add 180°: 307.2°. (d) BD is perpendicular to AC; \(DA=120\cos A=15\) m.

(a)(i) \(BC=\sqrt{8^2+13^2-2(8)(13)\cos79^\circ}=13.903658\ldots\approx13.9\) m. (ii) \(\frac{\sin C}{13}=\frac{\sin79^\circ}{BC}\Rightarrow C=66.6^\circ\).

(b)(i) \(\frac12(x+4)(4x-5)\sin30^\circ=70\), hence \((x+4)(4x-5)=280\Rightarrow4x^2+11x-20=280\Rightarrow4x^2+11x-300=0\). (ii) \(x=\frac{-11\pm\sqrt{11^2-4(4)(-300)}}{8}=\frac{-11\pm\sqrt{4921}}8\), so x=−10.14 or 7.39. (iii) Only the positive root gives positive side lengths; \(DE=x+4=11.39\) m, or 11.4 m to 3 significant figures.

(a) \(h=107\tan19^\circ=36.8\) m. (b) \(\cos\angle BAC=\frac{158^2+132^2-107^2}{2(158)(132)}\), giving 42.1°. (c) \(\frac{\sin\angle CAD}{86}=\frac{\sin116^\circ}{132}\), giving 35.8°. The obtuse alternative is impossible with the given 116°.

(d) In triangle ACD, \(\angle ACD=180-116-\angle CAD\approx28.2^\circ\) and \(AD=\frac{132\sin\angle ACD}{\sin116^\circ}\). Add the triangle areas \(\frac12(158)(132)\sin\angle BAC+\frac12(132)(AD)\sin\angle CAD\approx9670\) m², using unrounded angles. (e) Bearing of C from A=\(70-35.843=34.157\)°. Reverse bearing=\(34.157+180=214.2\)°.

\(\frac{\sin B}{5.9}=\frac{\sin84.6^\circ}{17.8}\Rightarrow B=\sin^{-1}\left(\frac{5.9\sin84.6^\circ}{17.8}\right)=19.268\ldots\approx19.3^\circ\). The obtuse alternative would make B+C exceed 180°.

\(\cos M=\frac{19^2+16^2-14^2}{2(19)(16)}=\frac{421}{608}\). Thus \(M=\cos^{-1}(421/608)=46.176\ldots\approx46.2^\circ\).

\(\cos x=\frac{11^2+5^2-13^2}{2(11)(5)}=-\frac{23}{110}\). \(x=\cos^{-1}(-23/110)=102.069\ldots\approx102.1^\circ\).

(a) \(OA=\sqrt{8^2+7^2-2(8)(7)\cos78^\circ}=9.471\ldots\approx9.47\) cm. (b) \(\sin A=\frac{7\sin78^\circ}{OA}\Rightarrow A\approx46.3^\circ\). (c) Arc CB=29.5−(8+7+OA)≈5.029 cm. \(\theta=\frac{\text{arc CB}}{2\pi(7)}360^\circ\approx41.2^\circ\). (d) \(\frac12(8)(7)\sin78^\circ+\frac{\theta}{360}\pi(7)^2\approx45.0\) cm², using unrounded OA and θ.

(a) \(\frac12(8.9)(12.5)\sin130.6^\circ=42.2\) cm². (b)(i) \(\sin C=\frac{11.6\sin123.5^\circ}{21.3}\) gives C≈27.0°. (ii) CBD=180−123.5−27.006=29.494°; ABD=90−29.494=60.506°. Cosine rule: \(AD=\sqrt{18^2+11.6^2-2(18)(11.6)\cos60.506^\circ}\approx15.9\) cm.

(a)864/(7/3)=370.286…≈370 km/h. (b) \(GB=\sqrt{864^2+928^2-2(864)(928)\cos67^\circ}\approx991\) km. (c)(i)133+180=313°. (ii)H→G bearing=133+67=200°, so G→H=020°. \(\angle HGB=\cos^{-1}\left(\frac{864^2+GB^2-928^2}{2(864)GB}\right)\approx59.58^\circ\). G→B bearing≈020°+59.58°=079.6°.

(a) \(\cos C=\frac{13^2+4^2-11^2}{2(13)(4)}=\frac8{13}\), so ACB≈52.02°. (b) \(\sin CAD=\frac{8\sin80^\circ}{13}\) gives CAD≈37.3°; obtuse alternative cannot coexist with 80°. ACD=180−80−CAD≈62.7°. (c) Add triangle areas: \(\frac12(13)(4)\sin52.02^\circ+\frac12(13)(8)\sin62.70^\circ\approx66.7\) cm².

(a) \(\frac12(2.8)(8.3)\sin79^\circ=11.4065\ldots\approx11.4\) cm². (b) Area factor \(4.5^2=20.25\), so area \(11.4065\ldots(20.25)=230.982\ldots\approx231\) cm².

(a)(i) Cosine rule: \(XY=\sqrt{12.5^2+9.9^2-2(12.5)(9.9)\cos23^\circ}=5.14\) cm. (ii) In TYZ, sine rule \(\frac{TZ}{\sin72^\circ}=\frac{9.9}{\sin37^\circ}\), so TZ≈15.6 cm. (b) The outer boundary is two radii plus three arcs: \(20.5=2r+3\frac{65}{360}(2\pi r)\). Thus \(r=\frac{20.5}{2+13\pi/12}=3.793913\ldots\approx3.79\) cm.

\(17=\frac12(8)(7)\sin\theta\Rightarrow\sin\theta=17/28\). Thus \(\theta=37.3832\ldots\approx37.4^\circ\) or \(180-37.3832\ldots=142.6168\ldots\approx142.6^\circ\).

Sine rule: \(\frac{AC}{\sin30^\circ}=\frac{8.15}{\sin110^\circ}\Rightarrow AC=4.33652\ldots\approx4.34\) m.

Sine rule: \(\frac{\sin x}{24}=\frac{\sin71.8^\circ}{39}\Rightarrow x=35.771\ldots\approx35.8^\circ\). The supplementary angle is impossible because it and 71.8° exceed 180°.

(a) \(AC=\sqrt{525^2+872^2-2(525)(872)\cos104^\circ}=1121.388732\ldots\approx1120\) m. (b) \(BQ=872\tan1^\circ=15.220817\ldots\) m; \(\theta=\tan^{-1}(BQ/525)=1.660656\ldots\approx1.7^\circ\).

(c)(i) \(\frac12(525)(872)\sin104^\circ=222100.691745\ldots\approx222000\) m². (ii) 1 map cm represents 200 m, so 1 map cm² represents 40000 m². Area=\(222100.691745\ldots/40000=5.552517\ldots\approx5.55\) cm².

Sine rule: \(\sin C=5.9\sin79^\circ/12.6\), so \(C\approx27.364578^\circ\). Then \(B=180-79-C\approx73.6^\circ\).

(a) \(12.5^2=x^2+8.5^2-2(8.5)x\cos60^\circ\) gives \(156.25=x^2+72.25-8.5x\), then multiply by 2. (b) \(x=\frac{17\pm\sqrt{17^2+4(2)(168)}}4=\frac{17\pm\sqrt{1633}}4\), giving −5.85 and 14.35. Use the positive root as a length.

(c) \(CD=x\sin46^\circ/\sin58^\circ\approx12.17\) cm. (d) Find \(AD=x\sin76^\circ/\sin58^\circ\). Total area=\(\frac12(8.5)x\sin60^\circ+\frac12x(AD)\sin46^\circ\approx137.6\) cm², or 138 cm² to 3 significant figures.

The cosine rule gives \(\cos p=\frac{2.8^2+3.6^2-5.3^2}{2(2.8)(3.6)}\). Therefore \(p=111.1989\ldots^\circ\approx111.2^\circ\).

\(A=\frac12(22.3)(27.6)\sin25^\circ=130.0565\ldots\approx130\) cm².

(a) \(\frac12(7)(10)\sin35^\circ=20.0752\ldots\approx20.1\) cm². (b) By the cosine rule, \(AC^2=7^2+10^2-2(7)(10)\cos35^\circ\). Hence \(AC=5.8582\ldots\approx5.86\) cm.

(a) The smaller angle AOB is \(360^\circ-210^\circ=150^\circ\). Since \(OA=OB\), the base angles are \((180^\circ-150^\circ)/2=15^\circ\). (b) The cosine rule gives \(AB^2=8^2+8^2-2(8)(8)\cos150^\circ\). Thus \(AB=15.4548\ldots\approx15.5\) cm. (c) The sine rule gives \(\frac{\sin B}{8}=\frac{\sin72^\circ}{AB}\). Hence \(B=29.4921\ldots^\circ\approx29.5^\circ\). The supplementary angle is impossible because it would exceed the remaining angle sum.

(d) Use the unrounded AB and B. The sector area is \(\frac{210}{360}\pi(8)^2\). Triangle AOB has area \(\frac12(8)(8)\sin150^\circ=16\). In triangle ABC, \(A=180^\circ-72^\circ-B=78.5079\ldots^\circ\), so its area is \(\frac12(8)(AB)\sin A\). Total area \(=193.8660\ldots\approx194\) cm². (e) The area scale factor is \(4^2=16\). Actual area \(=193.8660\ldots/16=12.1166\ldots\approx12.1\) cm².

(a) \(A=\frac12(6.2)(4.7)\sin82^\circ=14.4282\ldots\approx14.4\) cm². (b) \(2050=\frac12(107)(75)\sin x\), so \(\sin x=\frac{4100}{8025}=\frac{164}{321}\). The diagram gives an acute angle, so \(x\approx30.7^\circ\).

(a) \(A=\tfrac12(16)(5.4)\sin62^\circ=38.1433\ldots\approx38.1\) cm². (b) Let the angle opposite 6.7 cm be z. By the sine rule, \(\sin z/6.7=\sin48^\circ/8.4\), giving \(z\approx36.35^\circ\). Hence \(x=180^\circ-48^\circ-z\approx95.6^\circ\). (c) Included angle APB is \(146-63=83^\circ\). Cosine rule: \(AB=\sqrt{180^2+245^2-2(180)(245)\cos83^\circ}\approx286\) km.

\(\frac{BC}{\sin30^\circ}=\frac{24}{\sin100^\circ}\Rightarrow BC=\frac{24\sin30^\circ}{\sin100^\circ}\approx12.2\) cm.

Let angle QPR be θ. \(38.5=\tfrac12(10)(9)\sin\theta\Rightarrow\sin\theta=77/90\). The diagram shows the acute angle, giving \(\theta\approx58.82^\circ\). Apply the cosine rule using the unrounded angle: \(QR=\sqrt{10^2+9^2-2(10)(9)\cos\theta}\approx9.37\) cm.

(a)(i) \(PQ=\sqrt{7.6^2+8.4^2-2(7.6)(8.4)\cos62^\circ}\approx8.27\) cm. (ii) \(A=\tfrac12(7.6)(8.4)\sin62^\circ\approx28.2\) cm². (b) Angle HGJ \(=126-45=81^\circ\). The bearing of G from H is 225°, so angle GHJ \(=225-164=61^\circ\). By the sine rule, \(GJ=63\frac{\sin61^\circ}{\sin81^\circ}\approx55.8\) km.

(a) \(AB=\sqrt{95.5^2+83.1^2-2(95.5)(83.1)\cos101^\circ}=138.037\ldots\) km, which rounds to 138 km. (b) \(\sin\angle CAB=83.1\sin101^\circ/AB\), giving \(\angle CAB\approx36.224^\circ\). From the diagram B lies to the north of the ray AC, so the bearing is \(146-36.224\approx109.776^\circ\approx110^\circ\). (c) The angle between the north-west sailing direction and BL is \(45+21=66^\circ\). At closest approach the line to L is perpendicular to the route, so sailing distance \(46.2\cos66^\circ=18.791\ldots\approx18.8\) km.

By the sine rule, \(y/\sin39^\circ=12.4/\sin74^\circ\), so \(y=12.4\sin39^\circ/\sin74^\circ\approx8.12\) cm.

(a) Bisect BC to form a right triangle: \(AC=0.6/\cos68^\circ=1.6016\ldots\approx1.60\) m. (b) Cosine rule: \(\cos D=(1.9^2+2.3^2-AC^2)/(2(1.9)(2.3))\), so \(D\approx43.5^\circ\). (c) Height of BCD is \(\sqrt{2.3^2-0.6^2}\), giving \(A=\tfrac12(1.2)\sqrt{2.3^2-0.6^2}\approx1.33\) m². (d) \(\theta=\sin^{-1}(1.25/1.9)\approx41.1^\circ\).

(a) \(CD=\sqrt{2380^2+680^2-2(2380)(680)\cos65^\circ}\approx2180\) km. (b) \(\sin\angle KMC=2380\sin40^\circ/1560\), giving the acute angle 78.7°. (c) Bearing of C from M is \(50^\circ+78.712\ldots^\circ=128.712\ldots^\circ\). Reverse it by adding 180°: approximately 309°. (d)(i) 23 39. (ii) \(1560/2.4=650\) km/h.

The largest angle is opposite 9 cm. \(\cos\theta=(2^2+8^2-9^2)/(2(2)(8))=-13/32\), so \(\theta\approx114^\circ\).

Use the sine rule: \(PR/\sin37^\circ=12.5/\sin66^\circ\). Hence \(PR\approx8.23\) cm.

(a) \(AD=80\sin55^\circ/\sin49^\circ\approx86.8\) m. (b) \(\cos C=(95^2+90^2-80^2)/(2(95)(90))\), so C ≈ 51.2°. (c) Angle ADB = 76°. Add triangle areas: \(A=\tfrac12(AD)(80)\sin76^\circ+\tfrac12(95)(90)\sin C=6699.71\ldots\approx6700\) m². (d) Using the unrounded area, \(6699.71\ldots(3250/10000)=2177.406\ldots\approx2180\) dollars (3 significant figures).

(a)(i) \(97=\tfrac12(12)(17)\sin P\Rightarrow P=\sin^{-1}(194/204)=71.98\ldots\approx72.0^\circ\). (ii) PX = 6 cm; \(XR^2=6^2+17^2-2(6)(17)\cos P\), so XR ≈ 16.2 cm. (b) Shared height \(h=9.4\sin37^\circ=a\cos42^\circ\), giving \(a=9.4\sin37^\circ/\cos42^\circ=7.612\ldots\approx7.61\). (c) \(\cos40^\circ=\sin50^\circ\), so x = 50° or 130°.

\(85=\tfrac12(AC)^2\sin110^\circ\Rightarrow AC=\sqrt{170/\sin110^\circ}=13.450\ldots\approx13.5\) cm.

(a)(i) AXC = DXB (vertically opposite). ACX = DBX (angles subtending chord AD). Two equal angles prove similarity. (ii)(a) \(AX/DX=CX/BX\Rightarrow DX=3.2(12.5)/4=10\) cm. (b) \(AC=\sqrt{3.2^2+4^2-2(3.2)(4)\cos110^\circ}=5.9156\ldots\approx5.92\) cm. (c) BXD = AXC = 110°, so \(A=\tfrac12(12.5)(10)\sin110^\circ=58.7308\ldots\approx58.7\) cm². (b) \(AB=30/\tan31^\circ\). Then \(DC=AB\tan37^\circ-30=7.6231\ldots\approx7.62\) m.

(a) \(BC=15.7/\sin30^\circ=31.4\) cm. (b) \(\sin\angle AEC=15.7\sin52^\circ/16.5\), giving \(\angle AEC=48.5734\ldots^\circ\approx48.57^\circ\). (c)(i) \(\angle ACE=180-52-48.5734\ldots=79.4266\ldots^\circ\). Since ACB = 60°, \(\angle ECD=180-60-79.4266\ldots=40.5734\ldots^\circ\approx40.6^\circ\). (ii) \(DE=\sqrt{16.5^2+23.4^2-2(16.5)(23.4)\cos40.5734\ldots^\circ}\approx15.3\) cm. (d) \(AB=15.7/\tan30^\circ,\quad AE=16.5\sin79.4266\ldots^\circ/\sin52^\circ\). Add triangles ABC, ACE and CDE: \(A=\tfrac12(AB)(15.7)+\tfrac12(15.7)(AE)\sin52^\circ+\tfrac12(16.5)(23.4)\sin40.5734\ldots^\circ\approx466\) cm².

(a)(i) \(\cos\angle LMN=\frac{12^2+21^2-15^2}{2(12)(21)}=5/7\). Thus \(\angle LMN=44.4153\ldots^\circ\approx44.4^\circ\). (ii) \(A=\tfrac12(12)(21)\sin44.4153\ldots^\circ=88.1816\ldots\approx88.2\) cm². (b) PRQ = 55°. Sine rule: \(PR=6.4\sin82^\circ/\sin55^\circ=7.73692\ldots\approx7.74\) cm.

(a)(i) ACD = 27° (same chord AD). (ii) AOD = 54° (central angle twice circumference angle). (iii) AED = 153° (opposite angles in cyclic ABDE). (b)(i) \(KN=\sqrt{45^2+32^2-2(45)(32)\cos100^\circ}=59.5743\ldots\approx59.6\) cm. (ii) \(324=\tfrac12(LM)(32)\sin67^\circ\Rightarrow LM=648/(32\sin67^\circ)\approx22.0\) cm. (iii) XZ corresponds to LN: linear scale \(16/32=1/2\). Area \(324(1/2)^2=81\) cm².

(a) \(\sin A/4=\sin65^\circ/6\Rightarrow A=\sin^{-1}(4\sin65^\circ/6)=37.1717\ldots^\circ\approx37.2^\circ\). The supplementary value is impossible with C = 65°. (b) \(B=180-65-A\approx77.8^\circ\). Area \(\tfrac12(6)(4)\sin B\approx11.7\) cm².

(a) Cosine rule: \(AD=\sqrt{55^2+70^2-2(55)(70)\cos40^\circ}=45.0162\ldots\approx45.0\) m. (b) Alternate angles give BDC = 40°, hence DBC = 108°. Sine rule gives \(BC=70\sin40^\circ/\sin32^\circ=84.9094\ldots\approx84.9\) m. (c)(i) Add the areas of ABD and BCD: \(A=\tfrac12(55)(70)\sin40^\circ+\tfrac12(70)(84.9094\ldots)\sin108^\circ=4063.7439\ldots\approx4060\) m². (ii) \(A_{\rm plan}=A\times10^4/200^2=1015.936\ldots\approx1020\) cm². (d) The perpendicular height from A to BD is \(55\sin40^\circ=35.3533\ldots\approx35.4\) m.

(a) \(\angle DBC=\tan^{-1}(1.8/2.4)=36.8699\ldots\approx36.9^\circ\). (b)(i) \(BD=\sqrt{2.4^2+1.8^2}=\sqrt9=3\) m. (ii) \(\cos\angle ABD=(6.46^2+3^2-8.6^2)/(2\cdot6.46\cdot3)\), so ABD = 126.8189… ≈ 127°. (c) Add the two triangle areas, retaining the unrounded angle: \(V=4[\tfrac12(2.4)(1.8)+\tfrac12(6.46)(3)\sin126.8189\ldots^\circ]=39.6687\ldots\approx39.7\) m³.

(a) \(AC=62\sin122^\circ/\sin26^\circ=119.9417\ldots\approx119.9\) m. (b) Cosine rule: \(BC=\sqrt{55^2+AC^2-2(55)(AC)\cos65^\circ}=108.7895\ldots\approx109\) m. (c) \(A_{ACD}=\tfrac12(62)(AC)\sin32^\circ=1970.342\ldots\approx1970\) m². (d) Add triangle ABC: \(A_{ABC}=\tfrac12(55)(AC)\sin65^\circ=2989.3627\ldots\) m². Cost \(4.50(A_{ACD}+A_{ABC})=22318.6713\ldots\approx22300\) dollars (3 significant figures).

\(A=\tfrac12(9)(15)\sin28^\circ=31.6893\ldots\approx31.7\) cm².

Sine rule: \(\sin A/13=\sin23^\circ/6\Rightarrow\sin A=13\sin23^\circ/6\). The acute inverse-sine angle is \(57.84206\ldots^\circ\), so the required obtuse angle is \(180-57.84206\ldots=122.15794\ldots\approx122.2^\circ\).

(a)(i) \(360/5=72^\circ\). (ii) Each base angle of DOC is \((180-72)/2=54^\circ\). The adjacent isosceles triangle contributes another 54°, so DCB = 108°. (iii) Since CD is horizontal, angle BCY = 72°; in right triangle CBY, \(\angle CBY=180-90-72=18^\circ\). (b) Bisect DOC: \(CD=2(7)\sin36^\circ=8.22899\ldots\approx8.23\) cm.

(c)(i) \(A_{DOC}=\tfrac12(7)^2\sin72^\circ\approx23.3\) cm². (ii) \(A_P=5A_{DOC}\approx116.5\) cm². (iii) \(A_{sector}=(72/360)\pi(7)^2\approx30.8\) cm². (iv) The bottom is \(7\cos36^\circ\) below O, so \(XY=7+7\cos36^\circ\approx12.7\) cm. (d) Rectangle width \(WX=14\cos18^\circ\). Thus \(A_R=[14\cos18^\circ][7+7\cos36^\circ]\) and \(n=A_R/A_P\approx1.45\). Ratio 1:1.45.

(a)(i) Cosine rule: \(\cos A=\frac{32^2+64^2-43^2}{2(32)(64)}=3271/4096\), giving \(A=37.004905\ldots^\circ\approx37.0^\circ\). (ii) \(\tfrac12(32)(64)\sin A\approx616\) m². (b) Sine rule in ACD: \(\sin D=64\sin55^\circ/70\), so \(D\approx48.5^\circ\). The obtuse alternative would make the triangle’s angle sum exceed 180°. \(C=180-55-D\approx76.5^\circ\); \(AD=70\sin C/\sin55^\circ\approx83.1\) m. Perimeter \(32+43+70+AD\approx228\) m.

(a) \(130=\tfrac12(16)(25)\sin B\Rightarrow\sin B=0.65\). The acute angle shown is \(B=\sin^{-1}(0.65)=40.5416\ldots^\circ\approx40.5^\circ\). (b) Using the unrounded angle, \(AC=\sqrt{16^2+25^2-2(16)(25)\cos B}\approx16.5\) cm. (c) \(130=\tfrac12(25)h\Rightarrow h=260/25=10.4\) cm.

(a)(i) Cosine rule: \(\cos F=(6^2+14^2-12^2)/(2\times6\times14)=11/21\), so \(F\approx58.4^\circ\). (ii) \(A=\tfrac12(6)(14)\sin F\approx35.8\) cm². (b) Sine rule: \(\sin Q/12=\sin117^\circ/18\Rightarrow Q\approx36.4^\circ\). The obtuse inverse-sine alternative is impossible with P already obtuse.

(a) \(\cos A=(8^2+6^2-9^2)/(2\cdot8\cdot6)=19/96\), so \(A=78.584\ldots^\circ\approx78.6^\circ\). (b)(i) \(\angle BOC=2A\) and OM bisects angle BOC, so 78.6°. (ii) \(\sin A=(9/2)/r\Rightarrow r=9/(2\sin A)=4.590\ldots\approx4.59\) cm. (c) Triangle area \(\tfrac12(8)(6)\sin A=23.525\ldots\) cm²; percentage \(100\cdot23.525\ldots/(\pi r^2)\approx35.5\%\).

(a) Cosine rule: \(AC=\sqrt{120^2+95^2-2(120)(95)\cos77^\circ}=135.263\ldots\approx135\) m. (b) \(\sin B/AC=\sin26^\circ/79\). The acute inverse-sine value is about 48.6°, but B is obtuse: 131.4°. (c) \(A=180-26-B\approx22.6^\circ\). The perpendicular path has length \(79\sin A\approx30.4\) m. (d) Area \(\tfrac12(120)(95)\sin77^\circ\approx5553.9\) m²; divide by 180 to get about 30.85. Round down: 30 houses.

(a)(i) \(BC=\sqrt{3^2+5^2-2(3)(5)\cos45^\circ}=3.57586\ldots\approx3.58\) m. (ii) Sine rule gives \(\sin C/3=\sin45^\circ/BC\Rightarrow C\approx36.4^\circ\). (b)(i) In isosceles ACD, angles at A and D are 52°, so \(DCA=180-104=76^\circ\). (ii) \(A=\tfrac12(3)(5)\sin45^\circ+\tfrac12(5)(5)\sin76^\circ\approx17.4\) m². (c) \(\cos\angle PAB=AP/AB=2/3\), so \(\angle PAB\approx48.2^\circ\).

(a) \(\tfrac12(26)(24)\sin40^\circ\approx201\) cm². (b) Cosine rule: \(AD=\sqrt{26^2+24^2-2(26)(24)\cos40^\circ}\approx17.2\) cm. (c) Angle C = 110°. Sine rule: \(BC/\sin30^\circ=24/\sin110^\circ\Rightarrow BC\approx12.8\) cm. (d) Perpendicular distance \(BC\sin40^\circ\approx8.21\) cm.

(a) \(BC=\sqrt{3^2+5^2}=\sqrt{34}\approx5.83\) cm. (b) \(\cos ACD=(5^2+8^2-11^2)/(2(5)(8))=-0.4\), so \(ACD=\cos^{-1}(-0.4)\approx113.6^\circ\). (c) Add triangles: \(A=\tfrac12(3)(5)+\tfrac12(5)(8)\sin(ACD)\approx25.8\) cm².

(a)(i) \(PQ/19.5=11/16.5\Rightarrow PQ=13\) cm. (ii) \(BC=\sqrt{19.5^2-16.5^2}=\sqrt{108}\approx10.4\) cm. (iii) \(ABC=\sin^{-1}(16.5/19.5)\approx57.8^\circ\). (iv) Volume scale factor \(32^3\), capacity \(0.02(32^3)=655.36\) litres. (b)(i) Cosine rule: \(DG=\sqrt{105^2+67^2-2(105)(67)\cos143^\circ}\approx164\) m. (ii) \(DEF=180-70-32=78^\circ\). Sine rule: \(EF=105\sin70^\circ/\sin78^\circ\approx101\) m.

(a)(i) \(LC=\sqrt{5480^2+3300^2-2(5480)(3300)\cos165^\circ}=8709.5353\ldots\approx8710\) km. (ii) \(\sin CLD=3300\sin165^\circ/LC\), so 5.63°. (b) \(01{:}50+13{:}45+7{:}00=22{:}35\). (c) Direct time \(8710/800=10.8875\) hours. Difference \(13.75-10.8875=2.8625\) hours = 2 h 51.75 min, rounding to 2 h 52 min.

(a) \(\cos PQR=(4.5^2+4^2-7^2)/(2(4.5)(4))=-12.75/36\), so \(PQR\approx110.7424^\circ\approx110.7^\circ\). (b) \(RS=7\sin40^\circ/\sin85^\circ\approx4.5167\approx4.52\) km. (c) \(PRS=180-40-85=55^\circ\). Add triangles: \(A=\tfrac12(4)(4.5)\sin110.7^\circ+\tfrac12(7)(4.52)\sin55^\circ\approx21.4\) km².