Practice Exam 3 - Chapter 6
Name:
Instructions:
Give yourself 50 minutes to complete this practice exam.
Justify each answer.
Each answer has equal weight.
1. Determine whether the Law of Cosines is needed to solve the triangle.
(Answer either “Yes” or “No.”)
A = 72◦ , B = 50◦ , a = 10.2 : No (Why? Angle-Angle-Side)
2. Determine whether the Law of Cosines is needed to solve the triangle.
(Answer either “Yes” or “No.”)
a = 6.0, b = 4.2, c = 2.1 : Yes (Why? Side-Side-Side)
3. Determine whether the Law of Cosines is needed to solve the triangle.
(Answer either “Yes” or “No.”)
C = 107◦ , a = 34, b = 55 Yes (Why? Side-Angle-Side)
4. Solve the triangle. A = 20◦ , C = 40◦ , b = 6.0.
B = 180◦ − (20◦ + 40◦ ) = 180◦ − 60◦ = 120◦
a
b
c
=
=
:
sin A
sin B
sin C
a=
b
6.0
· sin A =
· sin(20◦ )
◦
sin B
sin(120 )
c=
6.0
b
· sin C =
· sin(40◦ )
◦
sin B
sin(120 )
5. A triangle has B = 60◦ , a = 5, c = 6. Use the Law of Cosines to determine the exact length of b.
cos(60◦ ) = 21 . Use the second “side-length” version:
1
b = 5 + 6 − 2(5)(6)
2
1
= 25 + 36 − 60
2
= 61 − 30
= 21
2
b=
√
2
2
21
6. A triangle has a = 4, b = 10, C = 100◦ . Find the area.
Area is 12 (base)(height):
A = 21 (4)(10 sin(100◦ )) = 20 sin(100◦ )
For #7, #8, and #9, use ~u = h2, 5i and ~v = h−10, 3i.
7. Find ~u + ~v .
h2 + (−10), 5 + 3i = h−8, 8i
8. Find ~u − ~v .
h2 − (−10), 5 − 3i = h12, 2i
9. Find −6~u − 3~v .
h(−6)(2) + (−3)(−10), (−6)(5) + (−3)(3)i = h−12 + 30, −30 − 9i
= h18, −39i
10. Find the component form of the vector ~v with the conditions that the
initial point of ~v is (−8, 3) and the terminal point of ~v is (4, −11).
~v = h4 − (−8), −11 − 3i = h12, −14i
11. Find a unit vector in the direction of ~u = h−9, 20i.
p
√
√
||~u|| = (−9)2 + 202 = 81 + 400 = 481
−9
20
û = √
,√
481 481
12. Find the angle between the vectors ~u = h−1, 5i and ~v = h3, −2i.
||~u|| =
√
1 + 25 =
√
26 and ||~v || =
√
9+4=
√
13
~u · ~v = (−1)(3) + (5)(−2) = −3 − 10 = −13
−13
~u · ~v
−13
cos(θ) =
=√ √ =√
||~u||||~v ||
26 13
338
13
θ = arccos −
338
13. Are the vectors ~u = h6, 10i and ~v = h−5, 3i orthogonal?
~u · ~v = (6)(−5) + (10)(3) = −30 + 30 = 0
Yes.
(Why? Their dot product is zero. This means cos(θ) = 0 and θ = 90◦ )
14. Find the projection of ~u = h4, 5i onto ~v = h1, 10i.
~u · ~v
proj~v ~u =
~v
||~v ||2
~u · ~v = 4(1) + 5(10) = 54
√
√
||~v || = 11 + 102 = 101
54
proj~v ~u =
h1, 10i =
101
54 540
,
101 101
15. Write the complex number z = −3 + 3i in trigonometric form.
p
√
r = (−3)2 + 32 = 18 .
tan(θ) = −1.
This is in the second quadrant, so the angle is 180◦ − 45◦ = 135◦ .
z=
√
18(cos(135◦ ) + i · sin(135◦ ))
16. Write the complex number z = 7 (cos(60◦ ) + i sin(60◦ )) in standard
form.
z = 7 (cos(60◦ ) + i sin(60◦ ))
√ !
1
3
=7
+i·
2
2
√
7
7 3
z = +i·
2
2
17. Find the indicated power of the complex number.
3
5π
5π
2 cos
+ i sin
6
6
Cube the radius √
and triple the angle.
23 = 8 and 3
√
2
2
5π
Also, cos 5π
=
and
sin
2
2
2 = 2
3
z =8
√
3
2
√
+i·
3
2
5π
6
=
5π
2 .
√
√
=4 3+i·4 3
18. Find the indicated power of the complex number.
(5 + 5i)−1
r=
√
52 + 52 =
√
50. tan(θ) = 1. This is first quadarant, so θ = 45◦ .
√
z −1 = [ 50 (cos(45◦ ) + i · sin(45◦ ))]−1
√
= ( 50)−1 (cos(−45◦ ) + i · sin(−45◦ ))
√
√ !
√ −1
2
2
= ( 50)
−i·
2
2
19. Find the third roots of 2i.
z = 2(cos(90◦ ) + i · sin(90◦ ))
z0 = 2(cos(90◦ ) + i · sin(90◦ ))
√
◦
◦
z1 = 3 2(cos 90 +360
+ i · sin
3
√
◦
◦
+ i · sin
z2 = 3 2(cos 90 +720
3
90◦ +360◦
)
3
◦
◦
90 +720
)
3
=
=
√
3
√
3
2(cos(150◦ ) + i · sin(150◦ ))
2(cos(270◦ ) + i · sin(270◦ ))
20. Find all solutions of the equation x4 − i = 0.
x4 = i
Solutions are the fourth roots if i = cos(90◦ ) + i · sin(90◦ ):
z0 = cos(90◦ ) + i · sin(90◦ )
◦
◦
z1 = cos 90 +360
+ i · sin
4
◦
◦
z2 = cos 90 +720
+ i · sin
4
◦
◦
z3 = cos 90 +1080
+ i · sin
4
90◦ +360◦
= cos(112.5◦ ) + i · sin(112.5◦ )
4
90◦ +720◦
= cos(202.5◦ ) + i · sin(202.5◦ )
4
90◦ +1080◦
= cos(292.5◦ ) + i · sin(292.5◦ )
4