Question

A vector is in standard position, with its terminal point in the second quadrant and an x-coordinate of –5. The vector has a magnitude of StartRoot 61 EndRoot.Complete the statements describing the vector. The y-coordinate of the vector to the nearest tenth is . The direction angle to the nearest whole number is °.

Answer 1

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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