a.
The y-coordinate of the vector with its terminal point in the second quadrant is 6.
The magnitude of the vector, r = √(x² + y²) where x = x-coorcinate of vector = -5 and y = y-coordinate of vector.
Since r = √61 and r = √(x² + y²)
Making y subject of the formula, we have
y = √(r² - x²)
Substituting the values of the variables into the equation, we have
y = √((√61)² - (-5)²)
y = √(61 - 25)
y = ±√36
y = ±6
Since r is in the second quadrant, its y-coordinate is positive.
So, y = 6
So, the y-coordinate of the vector with its terminal point in the second quadrant is 6.
b.
The direction angle of the vector with its terminal point in the second quadrant is 130°
The direction angle of the vector is gotten from tanФ = y/x
Subsstituting x and y into the equation, we have
tanФ = y/x
tanФ = 6/-5
tanФ = -1.2
tan(180° - Ф) = 1.2
Taking inverse tan of both sides, we have
180° - Ф = tan⁻¹(1.2)
180° - Ф = 50.2°
Ф = 180° - 50.2°
Ф = 129.8°
Ф ≅ 130° to the neares whole number
The direction angle of the vector with its terminal point in the second quadrant is 130°.
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