Question

Please help if you know !! Much appreciated

Answer 1

QUESTION 1

Recall the mnemonics; SOH

`\sin L =(Opposite)/(Hypotenuse)`

`\sin L =(12)/(13)`

QUESTION 2

Recall the mnemonics; CAH

`\cos L =(Adjacent)/(Hypotenuse)`

`\sin L =(5)/(13)`

QUESTION 3

Recall the mnemonics; TOA

`\tan M =(Opposite)/(Adjacent)`

`\tan M =(5)/(12)`

QUESTION 4

`\sin M =(Opposite)/(Hypotenuse)`

`\sin M =(5)/(13)`

QUESTION 5

a) From the Pythagoras Theorem,

`AC^2+AB^2=BC^2`

`AC^2+4^2=5^2`

`AC^2+16=25`

`AC^2=25-16`

`AC^2=9`

`AC=√(9)`

`AC=3yd`

b) Using the cosine ratio,

`\cos (m\angle B)=(4)/(5)`

Take the inverse cosine of both sides;

`m\angle B=\cos ^(-1)((4)/(5))`

`m\angle B=36.86`

`\therefore m\angle B=36.9\degree` to the nearest tenth.

c)
`m\angle B +m\angle C=90\degree`

`\Rightarrow 36.9\degree +m\angle C=90\degree`

`\Rightarrow m\angle C=90\degree-36.9\degree`

`\Rightarrow m\angle C=53.1\degree`

QUESTION 6

a) using the sine ratio,

`\sin(51\degree)=(DE)/(18)`

`DE=18\sin(51\degree)`

`DE=13.99`

`\therefore DE=14.0yd` to the nearest tenth.

b) Using the cosine ratio,

`\cos (51\degree)=(EF)/(18)`

`EF=18\cos (51\degree)`

`EF=11.3yd` to the nearest tenth.

c)
`m\angle D+ m\angle F=90\degree`

`\Rightarrow m\angle D + 51\degree=90\degree`

`\Rightarrow m\angle D=90\degree-51\degree`

`\Rightarrow m\angle D=39\degree`

QUESTION 7

a) Using the Pythagoras Theorem;

`lG^2+GH^2=lH^2`

`l15^2+GH^2=17^2`

`225+GH^2=289`

`GH^2=289-225`

`GH^2=64`

`GH=√(64)`

`GH=8km`

b) Using the sine ratio,

`\sin(m\angle H)=(15)/(17)`

`m\angle H=\sin^(-1)((15)/(17))`

`m\angle H=61.9\degree`

c)
`m\angle l + m\angle H=90\degree`

`m\angle l +61.9\degree=90\degree`

`m\angle l =90\degree-61.9\degree`

`m\angle l =28.1\degree`

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;

`XZ^2=XY^2+YZ^2`

`XZ^2=3^2+5^2`

`XZ^2=9+25`

`XZ^2=34`

`XZ=√(34)` units

QUESTION 10.

a) Using the tangent ratio;

`\tan(m\angle X)=(5)/(3)`

`m\angle X=\tan^[-1}((5)/(3))`

`m\angle X=59.0\degree`

b)
`m\angle Z+m\angle X=90\degree`

`m\angle Z+59.0\degree=90\degree`

`m\angle Z=90\degree-59.0\degree`

`m\angle Z=31.0\degree`

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;

`BC^2=DC^2+DB^2`

`BC^2=1^2+1^2`

`BC^2=1+1`

`BC^2=2`

`BC=√(2)`

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

`\sin 30\degree=(1)/(2)`

QUESTION 13

`\cos 30\degree=(√(3) )/(2)`

QUESTION 14

`\tan 30\degree =(√(3) )/(3)`

QUESTION 15

`\tan 45\degree =1`

QUESTION 16

`\cos 45\degree =(√(2) )/(2)`

QUESTION 17

`\tan 45\degree =1`

Check attachment for the rest