MPM 2DI Practice Test – Chapter 6
Name: _______________________
1) Determine the value of x. Provide a reason and show your calculations.
C
C
a)
b)
x
8
x
50
70
B
25
B
A
A
x
12
10
B
c) A
15
10
50
C
A
x
d)
e)
15
C
5
C
20
x
17
B
A
12
D
Angle ACB = Angle CDB
22
B
----------------------------------------------------------------------------------------------------------------------------- ------------------------------2) A ship wishes to travel from point C to a town at point A. Point A lies 100km N200W of point C. Unfortunately the
ship traveled N500W instead. The captain realized his mistake after he had traveled 200km and reached point B.
Determine how far the ship is from point A and what direction it needs to head to reach point A.
B
x
C
A
x
C
A
B
3) A tower has begun to fall over and now slants at a 170 angle from vertical. Boris is standing on the ground and is
25m from the base of the tower. The angle of elevation from Boris to the top of the tower is 210. How far is it from
Boris’s location to the top of the tower?
----------------------------------------------------------------------------------------------------------------------------- ------------------------------4) Boris runs one direction at 5m/s and Crackers runs a slightly different direction at 3.5 m/s. After 5 seconds they
are 20 m apart. Determine the number of degrees between their paths.
5) If the hour hand on Boris’s watch is 3cm long and the minute hand is 10cm long.
Determine how much the distance between the tips of the hands will change between 10:00 and 10:15.
Answers
1a) 6.5
b) 124.5o
c) 11.5
d) 86.60
e) 3.1
2) 123.94 km the direction is E16.21oS (if you did the question correctly you would have calculated an angle of 23.79o)
3) 23.97m
4) 52.62o
5) 3.64 cm
MPM 2DI Practice Test Chapter 6 – Answers
1) Determine the value of x. Provide a reason and show your calculations.
C
C
a)
b)
x
8
x
B
25
B
B
A
A
x
12
10
50
70
c) A
15
10
50
C
c  10  15  2(10)(15) cos 50
sin A sin 25

12
10
12 sin 25
sin A 
10
 12 sin 25 
A  sin 1 

 10 
A  30.5
sin 70 sin 50

8
x
x(sin 70)  8(sin 50)
8(sin 50)
sin 70
x  6.5
x
2
2
2
c  132.16........
c  11.5
C  180  25  30.5
C  124.5
C
d)
e)
x
5
A
C
20
15
x
17
B
A
12
D
Angle ACB = Angle CDB
22
B
22 2  15 2  17 2  2(15)(17) cos C
484  225  289  510 cos C
484  225  289
 cos C
 510
 30 
C  cos 1 

 510 
C  86.6
CALCULATE side CB
5
A
C
20
CB 2  5 2  12 2  2(5)(12) cos 20
CB 2  56.23........
CB  7.5
x
7.5
12
D
13.2
B
CALCULCATE angle B and ACB
sin B sin 20

5
7 .5
 5 sin 20 
B  sin 1 

 7. 5 
B  13.2
ACB  180  20  13.2
ACB  146.8
angle CDB also equals 146.8
sin 146.8 sin 13.2

7.5
x
x sin 146.8  7.5 sin 13.2
7.5 sin 13.2
x
sin 146.8
x  3.1
MPM 2DI Practice Test Chapter 6 – Answers
2) A ship wishes to travel from point C to a town at point A. Point A lies 100km N20 0W of point C. Unfortunately the
ship traveled N500W instead. The captain realized his mistake after he had traveled 200km. Determine how far
the ship is from point A and what direction it needs to head to reach point A.
c 2  200 2  100 2  2(200)(100) cos 30
x
B
c 2  15358.98....
A
c  123.93 km
100
20
200
sin B sin 30

100 123.93
 100 sin 30 
B  sin 1 

 123.93 
B  23.79
30
C
16.21
B
50
23.79
The ship must travel S16.21o E for 123.93km
123.93
A
100
200
20
30
40
C
3) A tower has begun to fall over and now slants at a 170 angle from vertical. Boris is standing on the ground and is
25m from the base of the tower. The angle of elevation from Boris to the top of the tower is 21 0. How far is it from
Boris’s location to the top of the tower?
sin 86 sin 73

25
x
25 sin 73
x
sin 86
x  23.97
C
86
x
17
73
21
A
B
25
4) Boris runs one direction at 5m/s and Crackers runs a slightly different direction at 3.5 m/s. After 5 seconds they
are 20 m apart. Determine the number of degrees between their paths.
BORIS : 5  5  25
CRACKERS : 5  3.5  17.5
20  17.5  25  2(17.5)( 25) cos A
400  306.25  625
 cos A
 875
 531.25 
A  cos 1 

 875 
A  52.62
2
2
2
C
20
17.5
B
25
?
A
MPM 2DI Practice Test Chapter 6 – Answers
5) If the hour hand on Boris’s watch is 3cm long and the minute hand is 10cm long.
Determine how much the distance between the tips of the hands will change between 10:00 and 10:15.
10 pm
10:15pm
Between 2 numbers on a clock there are 30o
At 10 o’clock the hour hand points to the 10 and the
minute hand to the 12. Therefore there are 60o between
the two hands.
Between 2 numbers on a clock there are 30o
At 10 o’clock the hour hand points ¼ of the way between
the 10 and the 11 and the minute hand to the 3.
Therefore there are 142.5o between the two hands.
(30 x 4 ¾ =142.5)
x
x
10
60
3
10
3
x 2  3 2  10 2  2(3)(10) cos 60
x  79
x  8.89
2
The increase in the distance is 12.51 – 8.89 = 3.64 cm
142.5
x 2  3 2  10 2  2(3)(10) cos142.5
x 2  156.6....
x  12.51