Angle 1 . . . . . 14 parts
Angle 2 . . . . . 5 parts
Angle 3 . . . . . 11 parts
Total . . . . . . . . 30 parts
But the total of the angles in ANY triangle is always 180 degrees.
SO ... 180 degrees / 30 parts = 6 degrees per part.
Angle 1 . . . . . 14 parts x (6° / part) = 84°
Angle 2 . . . . . . 5 parts x (6° / part) = 30°
Angle 3 . . . . . 11 parts x (6° / part) = 66°
Check: Total . . . . . . . . . . . . . . . . . . 180° yay !
<span>Two characteristics of regular, periodic waveforms are :
</span><span>1) Amplitude - It is the </span><span>the length and width of waves, such as sound
waves, as they move or vibrate. An example would be how
much a radio wave moving back and forth.
</span><span>
2) Frequency - It is the number of waves cycles per unit of time, passing a
point per unit time. It is usually measured in Hertz.</span>
Answer:
Explanation:
One of the easiest ways to work with vectors is to use their components
a) To find the magnitude let's use the Pythagorean theorem
A² = Ax² + Ay² + Az²
A = √ 3² + 4² + (-5)² = √ 50
A = 7.07
B = √ Bx² + By² + Bz²
B = √ (-2)² + 0 + 6² = √ 40
B = 6.32
b) to find these angles the most practical use the concept of cosine directors with the formulas
Vector A
cos α = X / A
cos β = y / A
cos γ = z / A
cos α = 3 / 7.07
α = cos⁻¹ 0.424
α = 64.9º
cos β = 4 / 7.07
β = cos⁻¹ 0.5658
β = 55.5º
Cos γ = -5 / 7.07
γ=Cos⁻¹ (-0.7079)
γ= 135º
Vector B
cos α = X / B
cos β = y / B
cos γ = z / B
cos α = 0
α = 90º
cos β = -2 / 6.32
β = 108.4º
cos γ = 6 / 6.32
γ = 18.3º
c) find A + B
R = A + B = (3 + 0) i ^ + (4-2) j ^ + (-5 +6) k ^
R = 3 i ^ + 2j ^ + 1 k ^
d) to find the angles we use the scalar product
cos θ = A.B / | A | | B |
A.B = 0 i ^ -8 j ^ -30 k ^
Cos θ = [R (8 2 + 30 2)] / 7.07 6.32
Cos θ = 31.05 / 44.68
θ = 46º
e) find A-B
R = A-B = (3-0) i ^ + (4- (2)) j ^ + (-5 - 6) k ^
R = 3 i ^ + 6j ^ -11 k ^