Exercise 6.4.4
| 
u | = 3, | v
| = 2
θ = 45°
180° - 45° = 135° is the angle of the triangle opposite the side representing the sum of the vectors.
The angle opposite | v | is the angle between the vector u and the vector u + v.
| u + v |2
= | u |2 + | v |2 – 2 | u || v
| cos 135° = 
So, | u + v | = 4.635
Let φ denote the angle between u and u + v . Then using the Law of Cosines
Exercise 6.4.3
| v | = 4, θ= 45°
v = | v | < cos45°, sin45° > = 
Exercise 6.4.2
v = < -2, 2 >, | v | = 2√2
v is in quadrant II and tan θ = - 1, so θ = 135˚. The direction vector is v / | v | = < -√2/2, √2/2 >.
3v = < -6, 6 > and – 2v = < 4, -4 >.
Exercise 6.4.1
The magnitude of the direction vector < sinθ, cosθ > equals the square root of the sum of the squares of the components. So the magnitude is 1.
Exercise 6.3.4
First, we must write – 16 i in trigonometric form. The magnitude is 1, and the direction angle is π. So the trigonometric form of – 16 i is 16 ( cos π + i sin π ).
There will be four fourth roots. The magnitude of each of the fourth roots will be the fourth root of the magnitude of – 16 i. So the magnitude of each of the fourth roots will be 2. The angle of the first of the four fourth roots will be one-fourth of the angle for – 16 i , so the angle of the first of the four fourth roots will be π / 4. The four fourth roots will lie evenly spaced around the unit circle, so they will lie an angular distance of π / 2 radians apart. So the four angles associated with the four fourth roots will be π / 4, 3 π / 4, 5 π / 4 and 7 π / 4. So the roots will be:
2 [ cos (π / 4 ) + i sin (π / 4 ) ] = √2 + √2 i
2 [ cos (3 π / 4 ) + i sin (3 π / 4 ) ] = - √2 + √2 i
2 [ cos (5 π / 4) + i sin (5 π / 4 ) ] = - √2 - √2 i
2 [ cos (7 π / 4 ) + i sin (7 π / 4 ) ] = √2 - √2 i
Exercise 6.3.3
First we must put ( 1 + i ) in trigonometric form. The magnitude is √2 and the angle is 45°. The seventh power of the complex number will have a magnitude which is the seventh power of √2, or 8√2. The angle of the seventh power will be seven times the angle 45°, or 315°. Thus, the seventh power of ( 1 + i ) is
8√2 ( cos 315° + i sin 315° ) = 8 – 8 i
Exercise 6.3.2
z = | z | ( cos t + i sin t )
So z2 = | z |2 [ cos ( t + t ) + i sin ( t + t ) ] = | z |2 [ cos ( 2 t ) + i sin ( 2 t ) ]
Exercise 6.3.1
The magnitude of z is found by taking the square root of the sum of the squares if its components, thus
| z | = 3√2
The angle is the standard angle with terminal side through the point ( 3, 3 ), or π / 4. Thus, the trigonometric form for the complex number 3 + 3 i is
3√2 [ cos (π / 4) + i sin (π / 4) ]
Exercise 6.2.4
s = ( 1 / 2 ) ( a + b + c ) = 8.5
Area = √ [ s ( s – a )( s – b )( s – c ) ] = 9.045
Exercise 6.2.3
Exercise 6.2.2
| AB |2 = 282 + 262 – 2 (28)(26)cos158°=2810.0
So, | AB | ≈ 53 feet
Exercise 6.2.1
A = 28°, b = 14, c = 10
Since we do not have a side-sine pair, we must use the Law of Cosines. This problem is of the SAS type: we are given two sides and the included angle (the angle between the two given sides).
First, we will solve for the side opposite the given angle :
Next we will use the Law of Sines to solve the angle opposite the smaller of the two given sides. It's important that we solve for the angle opposite the smaller of the two sides, since we know that it will not be obtuse. The angle opposite the largest side of a triangle may be an obtuse angle, that is, it may be greater than 90°. It is best to find obtuse angles last by subtracting the other two angles from 180°.
B = 180° - 28° - 42.2° = 109.8°
Exercise 6.1.7
Remember, the Law of Sines states that the diameter of the circumscribed circle for a triangle ABC equals any side of the triangle divided by the sine of the opposite angle. Since we know the side AB has length 120 feet and the angle opposite the side AB is 8°, then the diameter must be
Exercise 6.1.6
Exercise 6.1.5
a = 10, b = 4, B = 20°
There are two angles between 0° and 180° whose sine is 0.8550:
58.8° and 121.2°.
Since the larger of the two angles, when added to the given angle 20° is only 141.2°, which is still less than 180°, the larger angle is a possible solution. There will be two different solutions to this triangle.
Solution #1
A = 58.8°
C = 180° - 20° - 58.8° = 101.2°
Solution #2
A = 121.2°
C = 180° - 20° - 121.2° = 38.8°
Exercise 6.1.4
A = 60°, a = 2, b = 10
Since the sine of an angle cannot exceed 1, there can be no solution to this triangle.
Exercise 6.1.3
A = 24°, B = 80°, c = 20
At first glance it might appear that we do not have a side-sine pair. But we can easily get the angle C opposite the given side and then apply the Law of Sines. This problem is of the type SAA.
C = 180° - 24° - 80° = 76°
Exercise 6.1.2
(a) The sum of two sides of a triangle must be larger than the third. Bud b + c = 6 and a = 7. So there is no solution.
(b) The sum of two angles of a triangle must be smaller than 180°. But A + B = 183°, so there is no solution.
(c) The sum of B and C is greater than 180° so there is no solution.
(d) The sum of the three angles is greater than 180°, so there is no solution.
Exercise 6.1.1
There is such a thing as similar triangles: triangles of different sizes, yet which have congruent angles. For example, two equilateral triangles of different sizes still have all their angles equal to 60°. So just knowing the angles tells us nothing about the size of the sides.