revisions and completion of weeks 9 and 10

hey mike,

 

just refer to the message ive already sent you in respect to the work i need completed. ill attach the week 11 file here.

SEND:

W

o

rk for Submission

Show all your workings clearly.

1.
Two ships are observed from point O.

At a particular time their positions

A

and

B

are

as shown on the right.

16

0

m

230 m

A

D

C

B

120 m

N

The distance between the ships at this time is

A.
3.0 km

B.
3.2 km

C.
4.5 km

D.
9.7 km

o
E.
10.4 km

Solution:

Using the cosine rule:

a2 = b2 + c2 – 2bc cos(∠A)

= 42+62 – 2*6*4 cos 300

= 16+36 – 48cos300

= 52 – 41.569

AB2 = 10.4

AB = 3.2 KM

2.
The bearing of an aeroplane, X, from a control tower, T, is 055°. Another aeroplane, Y, is
due east of control tower T. The bearing of aeroplane X from aeroplane Y is 302°.

A
D
C
B
120 m
N

25



The size of the angle TXY is

A.
32°

B.
35°

C.
55°

D.
58°

E.
113°

75

0

25
0

5 km

345

0

C

M

H

Solution:

900 – 550 = 350

= 1130

3.
A hiker walks 4 km from A on a bearing of 30º to a point B,

then 6 km on a bearing of 330º to a point C. The distance AC

in km is

A

o
30
sin
4

B

22
6448cos120

+-
o

C

22
6448cos120

++
o

D
6 sin 60º

E

52

COSINE RULE:

a2 = b2 + c2 – 2bc cos(∠A)

AC2 = 62 + 42 – 2*4*6 cos 120

Answer = Srt (AC2 = 62 + 42 – 2*4*6 cos 120) = B

4. Ship A and Ship B can both be seen from the lighthouse. Ship A is 5 km from the lighthouse,

on a bearing of 028o. Ship B is 5 km from Ship A, on a bearing of 130o.

(a) Two angles, x and y, are shown in the diagram.

(i) Determine the size of the angle x in degrees.

1800 – 280 = 1520

(ii) Determine the size of the angle y in degrees.

= 28 + 50 = 780

(b) Determine the bearing of the lighthouse from

Ship A.

= 130 + 50 + 28 = 2080

(c) Determine the bearing of the lighthouse from

Ship B.

= 2300

5. Starting from the camp at C, Tim takes a bearing of a mountain at M and notes it to be 25°.

He then walks 5 km to the hut at H and takes a second bearingof the same mountain and it is 345°.

(a) Work out the angles in the triangle CHM. Prove that it is a right angled triangle.

= 750 + 150 = 900 = Right angled trinagle

(b) From the mountain at M:

(i) what is the bearing of the camp? (ii) what is the bearing of the hut?

180 +25 = 2050
180 -15 = 1650

(c) How far is it (i) from the camp to the mountain (ii) from the hut to the mountain?

from the camp to the mountain

Cos 50 = 5/H

H = 5/C0S 50
= 7.78KM

Part II: from the hut to the mountain?
Tan 50 = o/5
O = TAN 50 *5
= 5.96KM

(d) Tim walks back to camp from the hut. What bearing does he follow? 345 – 90 = 2550

6.
The base of a lighthouse D, is at the top of a cliff 168 metres above sea level. The angle of depression from D to a boat at C is 28o. The boat heads towards the base of the cliff, A, and stops at B. The distance AB is 128 metres.

(a)
What is the angle of depression from D to B, correct to the nearest minute?
(b)
How far did the boat travel from C to B, correct to the nearest metre?
7.
Genie Construction is building a new shopping plaza on a plot of land that is a trapezium
with the two parallel sides pointing north. The following is a diagram, which is not drawn to
scale, of the plot:

Part 1

(a) Find the area and the perimeter of the site ABCD.
Area = ½(a+b) × h
= 0.5(120+160) * 230
= 32200m2
Perimeter:
Tan