(a) \(\cos30^\circ=12/AC\). Thus \(AC=12/(\sqrt3/2)=24/\sqrt3=8\sqrt3\text{ cm}\).

(b) Drop a perpendicular from \(B\) to \(AC\). Its length is \(12\sin30^\circ=6\text{ cm}\).

The sloping side is the hypotenuse of a right triangle with legs \(21-15=6\) and 8. Its length is \(\sqrt{6^2+8^2}=10\text{ cm}\). The perimeter is \(21+15+8+10=54\text{ cm}\).

\(\tan60^\circ=9/x\). As \(\tan60^\circ=\sqrt3\), \(x=9/\sqrt3=9\sqrt3/3=\boxed{3\sqrt3\text{ cm}}\).

The line from the centre to the midpoint of a chord is perpendicular to the chord. Thus \(AM=7\text{ cm}\) and triangle OAM is right-angled at M. \(\tan\angle OAM=6/7\), giving \(\boxed{40.6^\circ}\).

\(\tan29^\circ=XY/12.4\). Therefore \(XY=12.4\tan29^\circ=6.873432\ldots\), giving \(\boxed{6.87\text{ cm}}\).

Apply Pythagoras to a face diagonal and then to the space diagonal: \(AG=\sqrt{18^2+14^2+16^2}=\sqrt{776}=27.856776\ldots\), giving \(\boxed{27.9\text{ cm}}\).

(a) \(h^2+22^2=25^2\), so \(h=\sqrt{141}=11.874342\ldots\). Thus \(\boxed{h=11.9}\).

(b) The depression angle equals the elevation angle at B. \(\cos\theta=22/25\), so \(\theta=28.357636\ldots^\circ\), giving \(\boxed{28.4^\circ}\).

(a) \(AB=18\cos60^\circ=18/2=9\). Thus \(BC=17-9=8\). In right triangle BCD, \(CD=\sqrt{6^2+8^2}=\sqrt{100}=\boxed{10\text{ cm}}\).

(b) \(BE=18\sin60^\circ=9\sqrt3\), so \(DE=9\sqrt3-6\). The perimeter is \(AC+CD+DE+EA=17+10+(9\sqrt3-6)+18=\boxed{39+9\sqrt3\text{ cm}}\).

\(\tan30^\circ=6/x=1/\sqrt3\). Thus \(x=6\sqrt3\), giving \(\boxed{x=6\sqrt3\text{ cm}}\).

(a) \(\boxed{\cos90^\circ=0}\).

(b) Let the shared side be s. In the lower triangle, \(\sin30^\circ=n/s=1/2\), so \(s=2n\). In the upper triangle, \(\cos30^\circ=s/x=\sqrt3/2\). Hence \(x=2n/(\sqrt3/2)=\boxed{4n/\sqrt3=4n\sqrt3/3}\).

The shortest distance is the perpendicular height from Q to PR. In the right triangle formed, \(h=PQ\sin42^\circ=12.5\sin42^\circ=8.364132\ldots\), so \(\boxed{h=8.36\text{ cm}}\).

Let the cube edge be s. The projection of AP onto the base is AC, with \(AC=s\sqrt2\). Since \(CP=s\), the angle \(\theta=\angle PAC\) satisfies \(\tan\theta=s/(s\sqrt2)=1/\sqrt2\). Hence \(\boxed{\theta=35.3^\circ}\).

The longest straight segment in a cuboid is its space diagonal: \(d=\sqrt{5^2+5^2+7^2}=\sqrt{99}\text{ cm}\). Since \(99<100\), \(\sqrt{99}<10\), so the 10 cm rod cannot fit completely inside.

(a) Pythagoras gives \(AC^2=11^2-8^2=57\), so \(\boxed{AC=\sqrt{57}\text{ cm}}\). (b) Adjacent/hypotenuse gives \(\boxed{k=8/11}\).

The vertical difference is \(12.6-1.5=11.1\text{ m}\). Thus \(\tan\theta=11.1/33\), giving \(\boxed{\theta=18.6^\circ}\).

G is 5 cm above the base. Its projection lies 7 cm across and 15 cm along the base from A, so the horizontal distance is \(\sqrt{7^2+15^2}=\sqrt{274}\). Thus \(\tan\theta=5/\sqrt{274}\), giving \(\boxed{\theta=16.8^\circ}\).

(a) Three-dimensional Pythagoras: \(EF^2+6.4^2+5.1^2=13^2\). Hence \(EF=\sqrt{102.03}=\boxed{10.1\text{ cm}}\). (b) EG is the projection of AG onto the base. In right triangle AEG, \(\sin\angle AGE=6.4/13\), giving \(\boxed{29.5^\circ}\).

CH is the projection of CE onto the base, with \(CH=\sqrt{4^2+9.1^2}\). The vertical height EH=6.5 cm. Thus \(\tan\angle ECH=6.5/\sqrt{4^2+9.1^2}\), giving \(\boxed{33.2^\circ}\).

\(\cos x=8.5/14\). Hence \(x=\cos^{-1}(8.5/14)=\boxed{52.6^\circ}\).

AC is the projection of AV onto the base. \(AC=\sqrt{15^2+5^2}=\sqrt{250}\). Height CV=4. Thus \(\tan\angle VAC=4/\sqrt{250}\), giving \(\boxed{14.2^\circ}\).

\(\cos37^\circ=AB/8\), so \(AB=8\cos37^\circ=\boxed{6.39\text{ cm}}\).

\(QS=18\sin28^\circ\), so \(QT=18\sin28^\circ-4\). In triangle QTR, \(\tan\angle QTR=\frac9{18\sin28^\circ-4}\). Thus \(\boxed{\angle QTR=63.7^\circ}\).

Drop a perpendicular from C to AB. Its length h satisfies \(\sin33.14^\circ=h/17.2\). Thus \(h=17.2\sin33.14^\circ=\boxed{9.40\text{ cm}}\).

The projection of PB on the base is DB. \(DB=\sqrt{12^2+5^2}=13\text{ cm}\). In right triangle PDB, \(\tan\angle PBD=4/13\), so \(\boxed{17.1^\circ}\).

The projection of \(PC\) onto the base is \(AC\). \(AC=\sqrt{5^2+7^2}=\sqrt{74}\).

\(\tan\angle PCA=\frac4{\sqrt{74}},\quad\angle PCA=24.9^\circ\).

(a) End area \(145/15\). \(\frac12(6.4)(5.7)\sin\theta=145/15\) gives \(\theta=32.003479\ldots^\circ\approx32^\circ\).

(b) Cosine rule, keeping the unrounded angle: \(EA=\sqrt{6.4^2+5.7^2-2(6.4)(5.7)\cos\theta}=3.4027575\ldots=3.40\text{ cm}\).

(c) If height is \(h\), \(\frac12(6.4)h=145/15\), so \(h=3.0208333\ldots=3.02\text{ cm}\).

(d) \(BF=\sqrt{15^2+5.7^2}\). The vertical rise is \(h\), so \(\alpha=\sin^{-1}\left(\frac{h}{\sqrt{15^2+5.7^2}}\right)=10.850967\ldots^\circ=10.9^\circ\).

(e) \(145\text{ cm}^3=145\times10^{-6}\text{ m}^3\). Mass \(=938(145\times10^{-6})\times1000=136.01\text{ g}\approx136\text{ g}\).

(a) Base diagonal \(AC^2=2x^2\). The right triangle \(ACG\) gives \(2x^2+20^2=28^2\Rightarrow x^2=192\). Therefore \(x=\sqrt{192}=13.9\text{ cm}\).

(b) The angle is \(\angle RKM\) and \(\sin\theta=MR/KR\). Minimise the numerator and maximise the denominator: \(\theta_{\rm lower}=\sin^{-1}(29.5/37.5)=51.9^\circ\).

The projection of \(BH\) onto the face is \(AH\). \(AH=\sqrt{4.5^2+9.2^2}\). In right triangle \(BAH\), \(\tan\theta=\frac{15.1}{\sqrt{4.5^2+9.2^2}}\). Thus \(\theta=55.9^\circ\).

(a) Let \(N\) be the vertical projection of \(M\) onto \(DC\). The horizontal differences from \(B\) to \(N\) are 7 and 5, so \(BN^2=7^2+5^2=74\). \(BM=\sqrt{74+8^2}=\sqrt{138}=11.7\text{ cm}\).

(b) \(\theta=\tan^{-1}(8/\sqrt{74})=42.9^\circ\).

Pythagoras \((2t+3)^2=t^2+5^2\Rightarrow3t^2+12t-16=0\).

\(t=\frac{-12\pm\sqrt{144+192}}6=\frac{-12\pm\sqrt{336}}6\). Keep the positive root \(t\approx1.055\). Then \(w=\tan^{-1}(t/5)=11.9^\circ\).

(a) Base diagonal \(AC=\sqrt{18.6^2+9^2}\). Angle \(\tan^{-1}\left(14.5/\sqrt{18.6^2+9^2}\right)=35.1^\circ\).

(b) \(AG=\sqrt{18.6^2+9^2+14.5^2}=25.243019\ldots\). Thus \(GM=30-AG=4.756981\ldots=4.8\) cm.

\(\sqrt{24^2+12^2+8^2}=\sqrt{784}=28\text{ cm}\).

(a) Put BC = x, AB = 2x. \((2x)^2+x^2+6^2=24^2\) gives \(5x^2=540\), so \(AB=2\sqrt{108}=20.8\text{ cm}\). (b) The projection of AG on the base is AC. In right triangle ACG, \(\sin\theta=6/24\). Hence \(\theta=14.5^\circ\).

(a) \(h=\sqrt{14^2-10^2}=\sqrt{96}=9.80\text{ cm}\). (b) \(14+10+\sqrt{96}=33.8\text{ cm}\).

\(MC=\frac12\sqrt{12^2+10^2}=\sqrt{61}\) cm. In right triangle VMC, \(\cos\theta=MC/VC=\sqrt{61}/14\). Hence \(\theta=56.1^\circ\).

Draw perpendicular BH to AC. In right triangle BHC, \(\sin65^\circ=12.8/BC\). Thus \(BC=12.8/\sin65^\circ=14.1\) cm.

The required angle is ECO. Half the base diagonal \(OC=\frac12\sqrt{5^2+5^2}=5/\sqrt2\) cm. \(\tan\theta=9/(5/\sqrt2)\), hence \(\theta=68.6^\circ\).

\(\cos x=7/12\quad\Rightarrow\quad x=\cos^{-1}(7/12)=54.3^\circ\).

(a) \(AC=\sqrt{12^2+5^2+4^2}=\sqrt{185}=13.6\) cm. (b) The angle is CAP: \(\sin\theta=4/\sqrt{185}\), giving \(\theta=17.1^\circ\).

(a) \(\frac12(8.5)(10.8)=45.9\text{ cm}^2\). (b) Hypotenuse \(\sqrt{8.5^2+10.8^2}=13.7437\ldots\) cm. Perimeter \(8.5+10.8+13.7437\ldots=33.0\text{ cm}\) (3 significant figures).

Height \(h=495/(20\cdot5.5)=4.5\) cm. The projection of AG on the base has length \(\sqrt{20^2+5.5^2}=\sqrt{430.25}\) cm. Hence \(\theta=\tan^{-1}\left(4.5/\sqrt{430.25}\right)=12.2^\circ\).

Use Pythagoras on the three relevant faces: \(AH=\sqrt{6^2+6^2}=\sqrt{72},\quad AF=\sqrt{8^2+6^2}=10,\quad HF=\sqrt{8^2+6^2}=10\) cm. Cosine rule in triangle AHF gives \(\cos\angle HAF=\frac{72+100-100}{2\sqrt{72}(10)}\). Therefore \(\angle HAF=64.9^\circ\).

\(AM=\frac12\sqrt{11^2+11^2}=\frac{11}{\sqrt2}\) cm. In right triangle AMV, \(\cos\theta=AM/AV=\frac{11}{18.6\sqrt2}\). Thus \(\theta=65.3^\circ\).

Other side \(\sqrt{13^2-12^2}=5\) cm. Area \(12(5)=60\) cm².

PT is vertical and TV is the projection of PV onto the base. In right triangle PTV, \(PT=17.2\sin43^\circ=11.7\text{ cm}\).

\(AC=\sqrt{8^2+8^2}=8\sqrt2,\quad AM=4\sqrt2\). In right triangle VMA, \(\tan\theta=\frac{10}{4\sqrt2}\Rightarrow\theta=60.5^\circ\).

\(x=12\sin35^\circ=6.8829\ldots\approx6.88\) cm.

\(AC=\sqrt{18^2+7^2}=\sqrt{373}\). In right triangle ACG, \(\tan\theta=\frac{12}{\sqrt{373}}\Rightarrow\theta=31.9^\circ\).

(a) \(\frac12(1.6)(2.8)=2.24\) m². (b) \(AC=\sqrt{1.6^2+2.8^2}=\sqrt{10.4}=3.22\) m.

\(AC=\sqrt{2.5^2+4.1^2}=\sqrt{23.06}=4.80208\ldots\approx4.80\) cm.

OM is perpendicular to AB and AM=6 cm. \(OA=\sqrt{6^2+4.5^2}=\sqrt{56.25}=7.5\) cm.

(a) \(x^2+(2x-3)^2=6^2\Rightarrow5x^2-12x-27=0\). (b) \(x=\frac{12\pm\sqrt{144+540}}{10}=\frac{12\pm\sqrt{684}}{10}\), so x=3.82 or −1.42. A length must be positive, so use x≈3.81534. (c) \(P=x+(2x-3)+6=3x+3\approx14.4\) m. (d) x is the shorter leg: \(\theta=\sin^{-1}(x/6)\approx39.5^\circ\).

(a)(i) \(\angle BAC=\tan^{-1}(6/12)=26.565\ldots^\circ\). \(\angle DAE=\sin^{-1}(6/12)=30^\circ\). With CAD=60°, BAE≈116.6°. (ii) \(AB=\sqrt{12^2+6^2}=\sqrt{180}\approx13.4\) cm. (iii) \(AE=\sqrt{12^2-6^2}=\sqrt{108}\approx10.4\) cm. (iv) Add three triangle areas: \(\frac12(12)(6)+\frac{\sqrt3}4(12)^2+\frac12(6)\sqrt{108}=129.53\ldots\approx130\) cm². (b)(i)3. (ii) Projection of AR on the face is BR=\(\sqrt{4^2+5^2}=\sqrt{41}\). \(\theta=\tan^{-1}(8/\sqrt{41})\approx51.3^\circ\).

Horizontal projection length=\(\sqrt{16.2^2+5.5^2}\). \(\theta=\tan^{-1}\left(\frac8{\sqrt{16.2^2+5.5^2}}\right)\approx25.1^\circ\).

Projection of AP is AC; \(AC=\sqrt{12^2+6^2}=\sqrt{180}\) cm. \(\theta=\tan^{-1}\left(\frac4{\sqrt{180}}\right)=16.60\ldots^\circ\approx16.6^\circ\).

(a)(i) Sum of interior angles \((6-2)180=720\)°, so each is 720/6=120°. (ii) The horizontal through A bisects angle BAF, so \(PQ=2x\sin60^\circ=x\sqrt3\) cm. (iii) The two horizontal extensions beyond FE each equal \((10-x)\sin30^\circ\). Thus \(PS=10+2(10-x)\frac12=20-x\) cm. (b) Equal sides require \(x\sqrt3=20-x\Rightarrow x=\frac{20}{1+\sqrt3}\). Square side \(\frac{20\sqrt3}{1+\sqrt3}=12.67949\ldots\approx12.7\) cm.

The projection of AG on the base is EG, so the angle is AGE. Right triangle AEG gives \(\sin\theta=5/13\Rightarrow\theta=22.6199\ldots\approx22.6^\circ\).

\(\sin x=4/7\Rightarrow x=34.8499\ldots\approx34.8^\circ\).

The projection of AG on the base is AC=\(26\sqrt2\) cm. The height CG=26 cm, so \(\tan\theta=26/(26\sqrt2)=1/\sqrt2\Rightarrow\theta=35.2644\ldots\approx35.3^\circ\).

(a) MN=AB/2=4, so \(\tan\angle PNM=9/4\Rightarrow\angle PNM=66.0375\ldots\approx66.0^\circ\). (b) \(BM=\sqrt{4^2+3^2}=5\) cm. (c) \(\tan\theta=9/5\Rightarrow\theta=60.9454\ldots\approx60.9^\circ\).

(d) \(PC=PB=\sqrt{9^2+5^2}=\sqrt{106}\). In PBC, \(\cos\angle BPC=\frac{106+106-6^2}{2(106)}=44/53\). Cosine rule in PBX gives \(BX^2=106+7.5^2-2\sqrt{106}(7.5)(44/53)\), so \(BX=5.834404\ldots\approx5.83\) cm.

\(MC=\sqrt{6^2+6^2}=6\sqrt2\) cm. The required angle is ECM, so \(\tan\theta=9/(6\sqrt2)\Rightarrow\theta=46.6861\ldots\approx46.7^\circ\).

\(AC=\sqrt{4^2+3^2}=5\) cm. Q is vertically above C, so the angle is QAC. \(\tan\theta=2/5\Rightarrow\theta=21.8014\ldots\approx21.8^\circ\).

(a)(i) Angle unit=\(180/(2+3+5)=18\)°, so angles=36°,54°,90°. (ii) Shortest side lies opposite 36°: \(12\sin36^\circ=7.05342\ldots\approx7.05\) cm. (b)(i) \(7.8(5/3)=13\) cm. (ii) \(\sin^{-1}(3/5)=36.8699\ldots\approx36.9^\circ\).

\(\cos\angle BAC=\frac{AB}{AC}=\frac{0.9}{3.5}\), so \(\angle BAC=\cos^{-1}(0.9/3.5)=75.099\ldots^\circ\approx75.1^\circ\).

(a) The base diagonal is \(CH=\sqrt{7^2+3^2}=\sqrt{58}\) cm. Then \(CE=\sqrt{CH^2+EH^2}=\sqrt{58+25}=\sqrt{83}=9.1104\ldots\approx9.11\) cm. (b) The projection of CE onto the base is CH. Thus \(\tan\theta=\frac{EH}{CH}=\frac5{\sqrt{58}}\) and \(\theta=33.286\ldots^\circ\approx33.3^\circ\).

(a) Base diagonal \(BD=\sqrt{8^2+8^2}=8\sqrt2\) cm. Then \(BS=\sqrt{BD^2+8^2}=8\sqrt3=13.8564\ldots\approx13.9\) cm. (b) \(\tan\angle SBD=\frac{SD}{BD}=\frac8{8\sqrt2}=\frac1{\sqrt2}\), so \(\angle SBD\approx35.3^\circ\).

The angle of elevation from C is 52°. \(\tan52^\circ=\frac{300}{d}\), so \(d=\frac{300}{\tan52^\circ}=234.3857\ldots\approx234\) m.

Horizontal diagonal \(AC=\sqrt{18^2+15^2}=\sqrt{549}\) m. Then \(\tan\theta=6/\sqrt{549}\), so \(\theta=\tan^{-1}(6/\sqrt{549})\approx14.4^\circ\).

\(\cos x=2/5\Rightarrow x=\cos^{-1}(2/5)\approx66.4^\circ\).

(a)(i) \(\theta=\tan^{-1}(55/294)\approx10.6^\circ\). (ii) Remaining distance \(d=55/\tan24.8^\circ\approx119.0\) m, so \(x=294-d\approx175\) m. (b)(i) The horizontal distance from base centre to a corner is \(\sqrt{2.4^2+1.5^2}\). Hence \(y=\sqrt{4^2+2.4^2+1.5^2}=\sqrt{24.01}=4.9\) m. (ii) \(\theta=\sin^{-1}(4/4.9)\approx54.7^\circ\).

(a) A tangent is perpendicular to the radius at contact: 90°. (b) \(\tan37^\circ=OP/11\Rightarrow OP=11\tan37^\circ=8.289\ldots\approx8.29\) cm.

\(x^2+5^2=8^2\Rightarrow x=\sqrt{39}=6.2449\ldots\approx6.24\) cm.

\(\sin x=2/5\Rightarrow x=\sin^{-1}(0.4)=23.578\ldots\approx23.6^\circ\).

(a)(i) \(BT=\sqrt{60^2-50^2}=\sqrt{1100}\). Hence \(\theta=\tan^{-1}(\sqrt{1100}/70)\approx25.4^\circ\). (ii) \(AC=\sqrt{50^2+70^2-2(50)(70)\cos130^\circ}\approx109\) m. (iii) \(A=\tfrac12(50)(70)\sin130^\circ\approx1340\) m². (b) By three-dimensional Pythagoras, \(XY=\sqrt{45^2+22^2+12^2}=\sqrt{2653}\approx51.5\) cm.

(a) MA = 4 cm. In right triangle PMA, \(PM=\sqrt{6^2-4^2}=\sqrt{20}\approx4.47\) cm. (b) \(\theta=\cos^{-1}(4/6)\approx48.2^\circ\).

\(AB=8\cos28^\circ=7.0635\ldots\approx7.06\) cm.

The horizontal projection is AC: \(AC=\sqrt{4^2+3^2}=5\) cm. Thus \(\theta=\cos^{-1}(5/12)\approx65.4^\circ\).

(a)(i) \(AD=\sqrt{4.7^2+2.6^2}=5.3712\ldots\approx5.37\) cm. (ii) The horizontal distance from B to C is 17 − 11 − 2.6 = 3.4 cm. \(\angle BCD=\tan^{-1}(4.7/3.4)=54.118\ldots^\circ\approx54.1^\circ\). (iii) \(A=\tfrac12(11+17)(4.7)=65.8\) cm². (b) Linear factor 9.4/4.7 = 2 gives area factor 4, so 263.2 cm².

The projection of CE onto the base is AE. \(AE=\sqrt{12^2+5^2}=13\) cm. Then \(\tan\theta=6/13\Rightarrow\theta=24.7751\ldots^\circ\approx24.8^\circ\).

(a) BD bisects AC at right angles, so AX = CX = 6 cm and \(XB=20(2/5)=8\) cm. \(\angle ABC=2\tan^{-1}(6/8)=73.7398\ldots^\circ\approx73.7^\circ\). (b) \(A=\tfrac12(12)(20)=120\) cm².

\(\sin15^\circ=h/628\Rightarrow h=628\sin15^\circ=162.5384\ldots\approx160\) m.

\(x=29\sin53.2^\circ=23.2212\ldots\approx23.2\) cm.

\(H=170+58.4\tan78.3^\circ=452.003\ldots\approx452\) m.

The base diagonal is \(8\sqrt2\) cm, so \(MB=4\sqrt2\) cm. (a) \(PB=\sqrt{5^2+(4\sqrt2)^2}=\sqrt{57}=7.5498\ldots\approx7.55\) cm. (b) The projection of PB onto the base is MB. \(\theta=\tan^{-1}[5/(4\sqrt2)]=41.4729\ldots\approx41.5^\circ\).

The projection of DP is DB, where \(DB=\sqrt{10^2+5^2}=\sqrt{125}\) cm. (a) \(DP=\sqrt{125+6^2}=\sqrt{161}=12.6886\ldots\approx12.7\) cm. (b) \(\theta=\tan^{-1}(6/\sqrt{125})=28.220512\ldots\approx28.2^\circ\).

\(C=\cos^{-1}(4/5)=36.8699\ldots^\circ;\ A=90-C\approx53.1^\circ\).

(a) Base diagonal \(EG=\sqrt{2^2+1.5^2}=2.5\) m; \(EC=\sqrt{2.5^2+1.7^2}=\sqrt{9.14}\approx3.02\) m. (b) \(\theta=\tan^{-1}(1.7/2.5)\approx34.2^\circ\). (c)(i) 2.95 m. (ii) Yes: even the upper bound 2.95 m is shorter than EC ≈ 3.023 m.

(a)(i) \(AP=\sqrt{12^2+4^2+3^2}=\sqrt{169}=13\) cm. (ii) Horizontal projection AC = \(\sqrt{12^2+4^2}\). Elevation \(\tan^{-1}[3/\sqrt{160}]\approx13.3^\circ\). (b)(i) \(PBC=\tan^{-1}(3/4)\approx36.9^\circ\). (ii) In BXC, angle XBC = PBC. Sine rule gives \(XC/\sin(PBC)=4/\sin120^\circ\). Since \(\sin(PBC)=3/5\), \(XC=4(3/5)/\sin120^\circ\approx2.77\) cm.

There are five horizontal pieces, two vertical pieces and one diagonal: \(L=5(3.5)+2(1.5)+\sqrt{3.5^2+1.5^2}=20.5+\sqrt{14.5}\approx24.3\) m.

(a) \(OC=6+4=10\) cm. (b) The radii to P and T are perpendicular to PT; their lengths differ by 2 cm. The right triangle has hypotenuse 10 and vertical side 2, so \(PT=\sqrt{10^2-2^2}=\sqrt{96}\approx9.80\) cm.

(a) \(h=730\sin37.1^\circ\approx440\) m. (b) \(730/3.65=200\) s = 3 min 20 s.

\(\tan17^\circ=PF/55\Rightarrow PF=55\tan17^\circ\approx16.8\) m.

(a) \(BD=\sqrt{10^2+10^2}=10\sqrt2\approx14.1\) cm. (b) \(MB=5\sqrt2\). In right triangle PMB, \(MP=\sqrt{8^2-(5\sqrt2)^2}=\sqrt{14}\approx3.74\) cm.

\(\sin JRG=20/50\Rightarrow JRG=\sin^{-1}(0.4)\approx23.6^\circ\).