QUESTION 1
Recall the mnemonics; SOH
[tex]\sin L =\frac{Opposite}{Hypotenuse}[/tex]
[tex]\sin L =\frac{12}{13}[/tex]
QUESTION 2
Recall the mnemonics; CAH
[tex]\cos L =\frac{Adjacent}{Hypotenuse}[/tex]
[tex]\sin L =\frac{5}{13}[/tex]
QUESTION 3
Recall 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 5
a) 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 6
a) 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 7
a) 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 8
We plot the points as shown in the diagram.
QUESTION 9
From 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] units
QUESTION 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 11
a) Triangle BCD is shown in the attachment.
The length of side DC=|3-2|=1 unit
The length of side DB=|4-3|=1 unit
Using 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
