QUESTION 1Recall the mnemonics; SOH[tex]\sin L =\frac{Opposite}{Hypotenuse}[/tex][tex]\sin L =\frac{12}{13}[/tex]QUESTION 2Recall the mnemonics; CAH[tex]\cos L =\frac{Adjacent}{Hypotenuse}[/tex][tex]\sin L =\frac{5}{13}[/tex]QUESTION 3Recall the mnemonics; TOA[tex]\tan M =\frac{Opposite}{Adjacent}[/tex][tex]\tan M =\frac{5}{12}[/tex]QUESTION 4[tex]\sin M =\frac{Opposite}{Hypotenuse}[/tex][tex]\sin M =\frac{5}{13}[/tex]QUESTION 5a) From the Pythagoras Theorem, [tex]AC^2+AB^2=BC^2[/tex][tex]AC^2+4^2=5^2[/tex][tex]AC^2+16=25[/tex][tex]AC^2=25-16[/tex][tex]AC^2=9[/tex][tex]AC=\sqrt{9}[/tex][tex]AC=3yd[/tex]b) Using the cosine ratio,[tex]\cos (m\angle B)=\frac{4}{5}[/tex]Take the inverse cosine of both sides;[tex] m\angle B=\cos ^{-1}(\frac{4}{5})[/tex][tex] m\angle B=36.86[/tex][tex]\therefore m\angle B=36.9\degree[/tex] to the nearest tenth.c) [tex]m\angle B +m\angle C=90\degree[/tex][tex]\Rightarrow 36.9\degree +m\angle C=90\degree[/tex][tex]\Rightarrow m\angle C=90\degree-36.9\degree[/tex][tex]\Rightarrow m\angle C=53.1\degree[/tex]QUESTION 6a) using the sine ratio,[tex]\sin(51\degree)=\frac{DE}{18}[/tex][tex]DE=18\sin(51\degree)[/tex][tex]DE=13.99[/tex][tex]\therefore DE=14.0yd[/tex] to the nearest tenth.b) Using the cosine ratio,[tex]\cos (51\degree)=\frac{EF}{18}[/tex][tex]EF=18\cos (51\degree)[/tex][tex]EF=11.3yd[/tex] to the nearest tenth.c) [tex]m\angle D+ m\angle F=90\degree[/tex] [tex]\Rightarrow m\angle D + 51\degree=90\degree[/tex] [tex]\Rightarrow m\angle D=90\degree-51\degree[/tex] [tex]\Rightarrow m\angle D=39\degree[/tex]QUESTION 7a) Using the Pythagoras Theorem;[tex]lG^2+GH^2=lH^2[/tex][tex]l15^2+GH^2=17^2[/tex][tex]225+GH^2=289[/tex][tex]GH^2=289-225[/tex][tex]GH^2=64[/tex][tex]GH=\sqrt{64}[/tex][tex]GH=8km[/tex]b) Using the sine ratio,[tex]\sin(m\angle H)=\frac{15}{17}[/tex][tex]m\angle H=\sin^{-1}(\frac{15}{17})[/tex][tex]m\angle H=61.9\degree[/tex]c) [tex]m\angle l + m\angle H=90\degree[/tex] [tex]m\angle l +61.9\degree=90\degree[/tex] [tex]m\angle l =90\degree-61.9\degree [/tex] [tex]m\angle l =28.1\degree [/tex]QUESTION 8We plot the points as shown in the diagram.QUESTION 9From the diagram, the side lengths XY and YZ can be obtained by counting the boxes. Each box is 1 unit.This implies that;XY =3 units YZ=5 units.We use Pythagoras Theorem, to obtain XZ.This implies that;[tex]XZ^2=XY^2+YZ^2[/tex][tex]XZ^2=3^2+5^2[/tex][tex]XZ^2=9+25[/tex][tex]XZ^2=34[/tex][tex]XZ=\sqrt{34}[/tex] unitsQUESTION 10.a) Using the tangent ratio;[tex]\tan(m\angle X)=\frac{5}{3}[/tex][tex]m\angle X=\tan^[-1}(\frac{5}{3})[/tex][tex]m\angle X=59.0\degree[/tex]b) [tex]m\angle Z+m\angle X=90\degree[/tex][tex]m\angle Z+59.0\degree=90\degree[/tex][tex]m\angle Z=90\degree-59.0\degree[/tex][tex]m\angle Z=31.0\degree[/tex]QUESTION 11a) Triangle BCD is shown in the attachment.The length of side DC=|3-2|=1 unitThe length of side DB=|4-3|=1 unitUsing Pythagoras Theorem;[tex]BC^2=DC^2+DB^2[/tex][tex]BC^2=1^2+1^2[/tex][tex]BC^2=1+1[/tex][tex]BC^2=2[/tex][tex]BC=\sqrt{2}[/tex]b) DB is perpendicular to DC, therefore m<D=90 degrees.The length of DB is equal the length of DC.This implies that;m<C=m<B=45 degrees.QUESTION 12[tex]\sin 30\degree=\frac{1}{2}[/tex]QUESTION 13[tex]\cos 30\degree=\frac{\sqrt{3} }{2}[/tex]QUESTION 14[tex]\tan 30\degree =\frac{\sqrt{3} }{3}[/tex]QUESTION 15[tex]\tan 45\degree =1[/tex]QUESTION 16[tex]\cos 45\degree =\frac{\sqrt{2} }{2}[/tex]QUESTION 17[tex]\tan 45\degree =1[/tex]Check attachment for the rest