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katen-ka-za [31]
3 years ago
12

Determine whether the vectors u and v are parallel, orthogonal, or neither.

Mathematics
1 answer:
Natalija [7]3 years ago
4 0
\mathbf u\cdot\mathbf v=\|\mathbf u\|\|\mathbf v\|\cos\theta

where \theta is the angle between the vectors. You have

10\times9+6\times5=\sqrt{10^2+6^2}\sqrt{9^2+5^2}\cos\theta\iff \cos\theta=\dfrac{30}{\sqrt{901}}\implies \theta\approx1.909^\circ

The vectors would be orthogonal if the dot product had been zero, but that's clearly not the case.

They would be parallel if the angle turned out to be 0^\circ or 180^\circ, but that's also not the case.

So the answer is neither.
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Step-by-step explanation:

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Please help with the question in the picture!
Stella [2.4K]

Answer:

Tan C = 3/4

Step-by-step explanation:

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∠ A = 90°, sin C = 3 / 5

<u>METHOD - I</u>

<u><em>Sin² C + Cos² C = 1</em></u>

Cos² C = 1 - Sin² C

Cos² C = 1 - \frac{9}{25}

Cos² C = \frac{25 - 9}{25}

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Cos C = \sqrt{\frac{16}{25} }

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As we know that

Tan C = \frac{Sin C}{Cos C }

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<em>Tan C = \frac{3}{4}</em>

<u>METHOD - II</u>

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therefore,  

AB ( Height ) = 3; BC ( Hypotenuse) = 5

<em>∵ ΔABC is Right triangle.</em>

<em>∴ By Pythagorean Theorem-</em>

<em>AB² + AC² = BC²</em>

<em>AC² </em><em>= </em><em>BC² </em><em>- </em><em> AB</em><em>² </em>

<em>AC² = 5² - 3²</em>

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<em>AC² = 16</em>

<em>AC  ( Base) = 4</em>

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<em>Tan C = \frac{Height}{Base}</em>

<em>Tan C = \frac{AB}{AC}</em>

<em>Hence Tan C = \frac{3}{4}</em>

<em />

4 0
3 years ago
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In the figure, m ll n. If the measure of angle 8 is (4x +7) and the measure of angle 2 is 107 degrees, what is the value of x? E
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Given:

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