Find cot theta. (-6,-3sqrt(5)

Mathematics

1 answer:

cupoosta [38]2 years ago

5 0

0 Pretend that is theta XD

cot 0 = $\frac{1}{tan0}$

Find tan 0.

tan 0 = $\frac{y}{x}$

tan 0 = $\frac{-3\sqrt{5}}{-6}$

tan 0 = $\frac{\sqrt{5}}{2}$

cot 0 = $\frac{1}{tan0}$

cot 0 = $\frac{1}{\frac{\sqrt{5} }{2} }$

Multiply 2/√5 to the numerator and the denominator (or multiply the reciprocal of the denominator to the top and bottom of the fraction)

cot 0 = $\frac{1}{\frac{\sqrt{5}}{2} } (\frac{\frac{2}{\sqrt{5}} }{\frac{2}{\sqrt{5}}} )$

cot 0 = $\frac{2}{\sqrt{5}}$

Multiply √5 to the numerator and the denominator

cot 0 = $\frac{2}{\sqrt{5}} (\frac{\sqrt{5} }{\sqrt{5} })$

cot 0 = $\frac{2\sqrt{5}}{5}$

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Find the distance points p (3,6) and q (7,3) to the nearest tenth?

Vladimir79 [104]

<h3>Answer: 5</h3>

=========================================================

One method is to plot the points P(3,6) and Q(7,3) on the same xy grid. Plot a third point R at (3,3). See the diagram below.

A right triangle forms in which we can find the legs PR = 3 and RQ = 4. The hypotenuse is found through the pythagorean theorem.

a^2+b^2=c^2

3^2+4^2 = c^2

9+16 = c^2

c^2 = 25

c = sqrt(25)

c = 5

This is the length of PQ

-----------------------------------

Or you can use the distance formula which is effectively using the pythagorean theorem just in a slightly different format (though it may not be obvious).

$d = \text{Distance from P to Q}\\\\d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(3-7)^2+(6-3)^2}\\\\d = \sqrt{(-4)^2+(3)^2}\\\\d = \sqrt{16+9}\\\\d = \sqrt{25}\\\\d = 5\\\\$