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Juli2301 [7.4K]
3 years ago
5

The distance between the two points pictured is d= Vn

Mathematics
1 answer:
katrin [286]3 years ago
5 0

the distance between points is:

d = 7.8 units

 d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)

 The ordered pairs are:

 (x1, y1) = (- 3, -2)

 (x2, y2) = (2,4)

 By applying the formula we have:

 d = root ((2 - (- 3)) ^ 2 + (4 - (- 2)) ^ 2)

 d = root (61)

 d = 7.8

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Given sin(u)= -7/25 and cos(v) = -4/5, what is the exact value of cos(u-v) if both angles are in quadrant 3
solmaris [256]

Given:

\sin (u)=-\dfrac{7}{25}

\cos (v)=-\dfrac{4}{5}

To find:

The exact value of cos(u-v) if both angles are in quadrant 3.

Solution:

In 3rd quadrant, cos and sin both trigonometric ratios are negative.

We have,

\sin (u)=-\dfrac{7}{25}

\cos (v)=-\dfrac{4}{5}

Now,

\cos (u)=-\sqrt{1-\sin^2 (u)}

\cos (u)=-\sqrt{1-(-\dfrac{7}{25})^2}

\cos (u)=-\sqrt{1-\dfrac{49}{625}}

\cos (u)=-\sqrt{\dfrac{625-49}{625}}

On further simplification, we get

\cos (u)=-\sqrt{\dfrac{576}{625}}

\cos (u)=-\dfrac{24}{25}

Similarly,

\sin (v)=-\sqrt{1-\cos^2 (v)}

\sin (v)=-\sqrt{1-(-\dfrac{4}{5})^2}

\sin (v)=-\sqrt{1-\dfrac{16}{25}}

\sin (v)=-\sqrt{\dfrac{25-16}{25}}

\sin (v)=-\sqrt{\dfrac{9}{25}}

\sin (v)=-\dfrac{3}{5}

Now,

\cos (u-v)=\cos u\cos v+\sin u\sin v

\cos (u-v)=\left(-\dfrac{24}{25}\right)\left(-\dfrac{4}{5}\right)+\left(-\dfrac{7}{25}\right)\left(-\dfrac{3}{25}\right)

\cos (u-v)=\dfrac{96}{625}+\dfrac{21}{625}

\cos (u-v)=\dfrac{1 17}{625}

Therefore, the value of cos (u-v) is 0.1872.

6 0
3 years ago
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